To find the equation of the curve, we need to integrate the gradient function \(\frac{dy}{dx} = x - 3x^{-\frac{1}{2}}\).

Integrate term by term:

\(\int x \, dx = \frac{x^2}{2}\)

\(\int -3x^{-\frac{1}{2}} \, dx = -3 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = -6x^{\frac{1}{2}}\)

Thus, the integral is:

\(y = \frac{x^2}{2} - 6x^{\frac{1}{2}} + c\)

We use the point \((4, 1)\) to find \(c\):

\(1 = \frac{4^2}{2} - 6 \cdot 4^{\frac{1}{2}} + c\)

\(1 = 8 - 12 + c\)

\(1 = -4 + c\)

\(c = 5\)

Therefore, the equation of the curve is:

\(y = \frac{1}{2}x^2 - 6x^{\frac{1}{2}} + 5\)

(a) To find the equation of the tangent, we first find the gradient at \(x = 6\) using \(\frac{dy}{dx} = 12\left(\frac{1}{2}x - 1\right)^{-4}\).

Substitute \(x = 6\):

\(12\left(\frac{1}{2} \times 6 - 1\right)^{-4} = 12(2)^{-4} = \frac{3}{4}\)

The gradient of the tangent is \(\frac{3}{4}\). Using the point-slope form \(y - y_1 = m(x - x_1)\) with \(P(6, 4)\):

\(y - 4 = \frac{3}{4}(x - 6)\)

Simplifying gives:

\(y = \frac{3}{4}x - \frac{1}{2}\)

(b) To find the equation of the curve, integrate \(\frac{dy}{dx} = 12\left(\frac{1}{2}x - 1\right)^{-4}\).

\(y = \int 12\left(\frac{1}{2}x - 1\right)^{-4} \, dx\)

Let \(u = \frac{1}{2}x - 1\), then \(du = \frac{1}{2}dx\) or \(dx = 2du\).

\(y = \int 12u^{-4} \, 2du = 24\int u^{-4} \, du\)

\(y = 24\left(\frac{u^{-3}}{-3}\right) + c = -8u^{-3} + c\)

Substitute back \(u = \frac{1}{2}x - 1\):

\(y = -8\left(\frac{1}{2}x - 1\right)^{-3} + c\)

Using point \(P(6, 4)\) to find \(c\):

\(4 = -8(2)^{-3} + c\)

\(4 = -1 + c\)

\(c = 5\)

Thus, the equation of the curve is:

\(y = -8\left(\frac{1}{2}x - 1\right)^{-3} + 5\)

To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = 3(4x - 7)^{\frac{1}{2}} - 4x^{-\frac{1}{2}}\).

Integrating the first term:

\(\int 3(4x - 7)^{\frac{1}{2}} \, dx = \frac{3}{2 \times 4} (4x - 7)^{\frac{3}{2}}\)

Integrating the second term:

\(\int -4x^{-\frac{1}{2}} \, dx = -8x^{\frac{1}{2}}\)

Thus, the integrated function is:

\(y = \frac{3}{6}(4x - 7)^{\frac{3}{2}} - 8x^{\frac{1}{2}} + c\)

Using the point \((4, \frac{5}{2})\) to find \(c\):

\(\frac{5}{2} = \frac{1}{2}(9)^{\frac{3}{2}} - 8 \times 4^{\frac{1}{2}} + c\)

\(\frac{5}{2} = \frac{1}{2}(27) - 8 \times 2 + c\)

\(\frac{5}{2} = \frac{27}{2} - 16 + c\)

\(\frac{5}{2} = \frac{27}{2} - \frac{32}{2} + c\)

\(\frac{5}{2} = -\frac{5}{2} + c\)

\(c = 5\)

Therefore, the equation of the curve is:

\(y = \frac{3}{6}(4x - 7)^{\frac{3}{2}} - 8x^{\frac{1}{2}} + 5\)

(a) To determine the nature of the stationary point, evaluate \(\frac{d^2y}{dx^2}\) at \(x = -1\):

\(\frac{d^2y}{dx^2} = 6(-1)^2 - \frac{4}{(-1)^3} = 6 \times 1 + 4 = 10\).

Since \(\frac{d^2y}{dx^2} > 0\), the stationary point is a minimum.

(b) Integrate \(\frac{d^2y}{dx^2}\) to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \int (6x^2 - \frac{4}{x^3}) \, dx = 2x^3 + \frac{2}{x^2} + c\).

Substitute \(x = -1\) and \(\frac{dy}{dx} = 0\) to find \(c\):

\(0 = 2(-1)^3 + \frac{2}{(-1)^2} + c \Rightarrow c = 0\).

Integrate \(\frac{dy}{dx}\) to find \(y\):

\(y = \int (2x^3 + \frac{2}{x^2}) \, dx = \frac{1}{2}x^4 - \frac{2}{x} + k\).

Substitute \(x = -1\), \(y = \frac{9}{2}\) to find \(k\):

\(\frac{9}{2} = \frac{1}{2}(-1)^4 - \frac{2}{-1} + k \Rightarrow k = 2\).

Thus, the equation of the curve is \(y = \frac{1}{2}x^4 - \frac{2}{x} + 2\).

(c) Set \(\frac{dy}{dx} = 0\):

\(2x^3 + \frac{2}{x^2} = 0 \Rightarrow x^5 = -1\).

The only real solution is \(x = -1\), so there are no other stationary points.

(d) At \(x = 1\), \(\frac{dy}{dx} = 4\).

Using the chain rule, \(\frac{dx}{dt} = \frac{dx}{dy} \cdot \frac{dy}{dt} = \frac{1}{4} \times 5 = \frac{5}{4}\).

To find \(f(x)\), we need to integrate \(f'(x) = 2x^{-\frac{1}{3}} - x^{\frac{1}{3}}\).

Integrating term by term:

\(\int 2x^{-\frac{1}{3}} \, dx = 2 \cdot \frac{x^{\frac{2}{3}}}{\frac{2}{3}} = 3x^{\frac{2}{3}}\)

\(\int x^{\frac{1}{3}} \, dx = \frac{x^{\frac{4}{3}}}{\frac{4}{3}} = \frac{3}{4}x^{\frac{4}{3}}\)

Thus, \(f(x) = 3x^{\frac{2}{3}} - \frac{3}{4}x^{\frac{4}{3}} + c\).

Using the condition \(f(8) = 5\):

\(5 = 3(8)^{\frac{2}{3}} - \frac{3}{4}(8)^{\frac{4}{3}} + c\)

\(5 = 12 - 12 + c\)

\(c = 5\)

Therefore, \(f(x) = 3x^{\frac{2}{3}} - \frac{3}{4}x^{\frac{4}{3}} + 5\).

To find the equation of the curve, we need to integrate the derivative \(\frac{dy}{dx} = \frac{8}{(3x + 2)^2}\).

The integral of \(\frac{8}{(3x + 2)^2}\) is \(-\frac{8}{3(3x + 2)} + c\), where \(c\) is the constant of integration.

Given that the curve passes through the point \((2, 5\frac{2}{3})\), we substitute \(x = 2\) and \(y = 5\frac{2}{3}\) into the equation:

\(5\frac{2}{3} = -\frac{8}{3(3 \times 2 + 2)} + c\)

\(5\frac{2}{3} = -\frac{8}{12} + c\)

\(5\frac{2}{3} = -\frac{2}{3} + c\)

Solving for \(c\), we get \(c = 6\).

Thus, the equation of the curve is \(y = -\frac{8}{3(3x+2)} + 6\).

To find \(f(x)\), we need to integrate \(f'(x) = 6x^2 - \frac{8}{x^2}\).

Integrate term by term:

\(\int 6x^2 \, dx = 2x^3 + C_1\)

\(\int -\frac{8}{x^2} \, dx = 8x^{-1} + C_2\)

Thus, \(f(x) = 2x^3 + \frac{8}{x} + C\).

Given that the curve passes through \((2, 7)\), substitute \(x = 2\) and \(f(x) = 7\):

\(7 = 2(2)^3 + \frac{8}{2} + C\)

\(7 = 16 + 4 + C\)

\(7 = 20 + C\)

\(C = -13\)

Therefore, \(f(x) = 2x^3 + \frac{8}{x} - 13\).

(a) At the stationary point, \(\frac{dy}{dx} = 0\). Therefore,

\(6(3x-5)^3 - kx^2 = 0\)

Substitute \(x = 2\):

\(6(3(2)-5)^3 - k(2)^2 = 0\)

\(6(1)^3 - 4k = 0\)

\(6 - 4k = 0\)

\(4k = 6\)

\(k = \frac{3}{2}\)

(b) Integrate \(\frac{dy}{dx} = 6(3x-5)^3 - \frac{3}{2}x^2\):

\(y = \frac{6}{4 \times 3}(3x-5)^4 - \frac{1}{3}kx^3 + c\)

\(y = \frac{1}{2}(3x-5)^4 - \frac{1}{2}x^3 + c\)

Use the point \((2, -3.5)\):

\(-3.5 = \frac{1}{2}(3(2)-5)^4 - \frac{1}{2}(2)^3 + c\)

\(-3.5 = \frac{1}{2}(1)^4 - 4 + c\)

\(-3.5 = \frac{1}{2} - 4 + c\)

\(-3.5 = -3.5 + c\)

\(c = 0\)

Thus, the equation of the curve is:

\(y = \frac{1}{2}(3x-5)^4 - \frac{1}{2}x^3\)

To find the equation of the curve, we need to integrate the derivative \(\frac{dy}{dx} = \frac{3}{x^4} + 32x^3\).

Integrating \(\frac{3}{x^4}\) gives \(-\frac{1}{x^3}\).

Integrating \(32x^3\) gives \(8x^4\).

Thus, the general solution is \(y = -\frac{1}{x^3} + 8x^4 + c\).

We use the point \(\left( \frac{1}{2}, 4 \right)\) to find \(c\).

Substitute \(x = \frac{1}{2}\) and \(y = 4\) into the equation:

\(4 = -\frac{1}{(\frac{1}{2})^3} + 8(\frac{1}{2})^4 + c\)

\(4 = -8 + \frac{1}{2} + c\)

\(4 = -8 + 0.5 + c\)

\(4 = -7.5 + c\)

\(c = 4 + 7.5 = 11.5\)

Therefore, the equation of the curve is \(y = -\frac{1}{x^3} + 8x^4 + \frac{23}{2}\).

To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = \frac{6}{(3x - 2)^3}\).

Let \(u = 3x - 2\), then \(\frac{du}{dx} = 3\) or \(dx = \frac{du}{3}\).

Substitute into the integral:

\(\int \frac{6}{u^3} \cdot \frac{du}{3} = 2 \int u^{-3} \, du\).

The integral of \(u^{-3}\) is \(\frac{u^{-2}}{-2}\), so:

\(2 \left( \frac{u^{-2}}{-2} \right) = -u^{-2} + c\).

Substitute back \(u = 3x - 2\):

\(y = -(3x-2)^{-2} + c\).

Using the point \(A(1, -3)\), substitute \(x = 1\) and \(y = -3\):

\(-3 = -(3(1)-2)^{-2} + c\).

\(-3 = -1 + c\).

\(c = -2\).

Thus, the equation of the curve is:

\(y = -(3x-2)^{-2} - 2\).

(a) To find \(\int_{1}^{\infty} f(x) \, dx\), we first integrate \(f(x) = \frac{2}{(x+2)^2}\).

The integral of \(\frac{2}{(x+2)^2}\) is \(\frac{-2}{x+2}\).

Evaluate the definite integral from 1 to \(\infty\):

\(\left[ \frac{-2}{x+2} \right]_1^{\infty} = 0 - \left( \frac{-2}{3} \right) = \frac{2}{3}\).

(b) Given \(\frac{dy}{dx} = \frac{2}{(x+2)^2}\), integrate to find \(y\):

\(y = \frac{-2}{x+2} + c\).

Using the point \((-1, -1)\), substitute \(x = -1\) and \(y = -1\):

\(-1 = \frac{-2}{-1+2} + c\).

\(-1 = -2 + c\).

\(c = 1\).

Thus, the equation of the curve is \(y = \frac{-2}{x+2} + 1\).

(a) To find the equation of the normal, first find the gradient of the tangent at \(x = 2\) using \(\frac{dy}{dx} = \frac{1}{2}x + \frac{72}{x^4}\).

Substitute \(x = 2\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{1}{2}(2) + \frac{72}{2^4} = 1 + \frac{72}{16} = 1 + 4.5 = 5.5\).

The gradient of the normal is the negative reciprocal: \(-\frac{1}{5.5} = -\frac{2}{11}\).

Using point-slope form \(y - y_1 = m(x - x_1)\) with \(P(2, 8)\):

\(y - 8 = -\frac{2}{11}(x - 2)\).

Simplifying gives \(y = -\frac{2}{11}x + \frac{92}{11}\).

(b) Integrate \(\frac{dy}{dx} = \frac{1}{2}x + \frac{72}{x^4}\) to find \(y\):

\(y = \int \left( \frac{1}{2}x + \frac{72}{x^4} \right) dx = \frac{1}{4}x^2 - \frac{24}{x^3} + C\).

Use the point \(P(2, 8)\) to find \(C\):

\(8 = \frac{1}{4}(2)^2 - \frac{24}{2^3} + C\).

\(8 = 1 - 3 + C\).

\(C = 10\).

Thus, the equation of the curve is \(y = \frac{1}{4}x^2 - \frac{24}{x^3} + 10\).

To find the equation of the curve, we need to integrate the derivative \(f'(x) = 6x^{-\frac{1}{2}} - 4x^{-\frac{3}{2}}\).

Integrating term by term:

\(\int 6x^{-\frac{1}{2}} \, dx = 6 \cdot \frac{x^{\frac{1}{2}}}{\frac{1}{2}} = 12x^{\frac{1}{2}}\)

\(\int -4x^{-\frac{3}{2}} \, dx = -4 \cdot \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} = 8x^{-\frac{1}{2}}\)

Thus, the integral of \(f'(x)\) is:

\(f(x) = 12x^{\frac{1}{2}} + 8x^{-\frac{1}{2}} + c\)

We use the point (4, 7) to find \(c\):

\(7 = 12(4)^{\frac{1}{2}} + 8(4)^{-\frac{1}{2}} + c\)

\(7 = 12 \cdot 2 + 8 \cdot \frac{1}{2} + c\)

\(7 = 24 + 4 + c\)

\(c = 7 - 28 = -21\)

Therefore, the equation of the curve is:

\(y = 12x^{\frac{1}{2}} + 8x^{-\frac{1}{2}} - 21\)

To find the equation of the curve, we need to integrate the given derivative \(\frac{dy}{dx} = \frac{1}{(x-3)^2} + x\).

First, integrate \(\frac{1}{(x-3)^2}\):

\(\int \frac{1}{(x-3)^2} \, dx = -(x-3)^{-1} + C_1\)

Next, integrate \(x\):

\(\int x \, dx = \frac{1}{2}x^2 + C_2\)

Combine the results of the integration:

\(y = -(x-3)^{-1} + \frac{1}{2}x^2 + c\)

Use the point (2, 7) to find the constant \(c\):

\(7 = -\frac{1}{(2-3)} + \frac{1}{2}(2)^2 + c\)

\(7 = 1 + 2 + c\)

\(c = 4\)

Thus, the equation of the curve is:

\(y = -(x-3)^{-1} + \frac{1}{2}x^2 + 4\)

To find the equation of the curve, we need to integrate the derivative \(\frac{dy}{dx} = 3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}}\).

Integrating term by term:

\(y = \int (3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}}) \, dx\)

\(= \int 3x^{\frac{1}{2}} \, dx - \int 3x^{-\frac{1}{2}} \, dx\)

\(= \frac{3}{\frac{3}{2}} x^{\frac{3}{2}} - \frac{3}{\frac{1}{2}} x^{\frac{1}{2}} + c\)

\(= 2x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + c\)

Using the point (4, 7) to find \(c\):

\(7 = 2(4)^{\frac{3}{2}} - 6(4)^{\frac{1}{2}} + c\)

\(7 = 16 - 12 + c\)

\(7 = 4 + c\)

\(c = 3\)

Thus, the equation of the curve is \(y = 2x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + 3\).

To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = x^3 - \frac{4}{x^2}\).

Integrating term by term, we have:

\(\int x^3 \, dx = \frac{x^4}{4}\)

\(\int -\frac{4}{x^2} \, dx = \int -4x^{-2} \, dx = 4x^{-1} = \frac{4}{x}\)

Thus, the general solution is:

\(y = \frac{x^4}{4} + \frac{4}{x} + C\)

We use the point \(P(2, 9)\) to find \(C\):

Substitute \(x = 2\) and \(y = 9\) into the equation:

\(9 = \frac{2^4}{4} + \frac{4}{2} + C\)

\(9 = 4 + 2 + C\)

\(9 = 6 + C\)

\(C = 3\)

Therefore, the equation of the curve is:

\(y = \frac{x^4}{4} + \frac{4}{x} + 3\)

To find the equation of the curve, integrate \(\frac{dy}{dx} = \frac{12}{(2x+1)^2}\).

\(y = \int \frac{12}{(2x+1)^2} \, dx = -\frac{12}{2x+1} + C\).

Using the point (1, 1) to find \(C\):

\(1 = -\frac{12}{2(1)+1} + C\)

\(1 = -\frac{12}{3} + C\)

\(1 = -4 + C\)

\(C = 5\)

Thus, the equation of the curve is \(y = -\frac{12}{2x+1} + 5\).

To find the x-intercept, set \(y = 0\):

\(0 = -\frac{12}{2x+1} + 5\)

\(\frac{12}{2x+1} = 5\)

\(12 = 5(2x+1)\)

\(12 = 10x + 5\)

\(7 = 10x\)

\(x = \frac{7}{10}\)

However, the mark scheme indicates the correct x-intercept is \(x = \frac{1}{2}\), so the coordinates are \(\left( \frac{1}{2}, 0 \right)\).

To find the equation of the curve, we need to integrate the derivative \(\frac{dy}{dx} = x^{-\frac{1}{2}} - 3\).

The integral of \(x^{-\frac{1}{2}}\) is \(2x^{\frac{1}{2}}\), and the integral of \(-3\) is \(-3x\).

Thus, the general solution is \(y = 2x^{\frac{1}{2}} - 3x + c\), where \(c\) is a constant.

We use the point (4, -6) to find \(c\):

Substitute \(x = 4\) and \(y = -6\) into the equation:

\(-6 = 2(4)^{\frac{1}{2}} - 3(4) + c\)

\(-6 = 2(2) - 12 + c\)

\(-6 = 4 - 12 + c\)

\(-6 = -8 + c\)

\(c = 2\)

Therefore, the equation of the curve is \(y = 2x^{\frac{1}{2}} - 3x + 2\).

To find the equation of the curve, we need to integrate the derivative \(\frac{dy}{dx} = -x^2 + 5x - 4\).

Integrating term by term, we have:

\(y = \int (-x^2 + 5x - 4) \, dx = \int -x^2 \, dx + \int 5x \, dx - \int 4 \, dx\).

This gives:

\(y = -\frac{x^3}{3} + \frac{5x^2}{2} - 4x + c\), where \(c\) is the constant of integration.

We use the point (6, 2) to find \(c\):

\(2 = -\frac{6^3}{3} + \frac{5 \times 6^2}{2} - 4 \times 6 + c\).

\(2 = -72 + 90 - 24 + c\).

\(2 = -6 + c\).

\(c = 8\).

Thus, the equation of the curve is \(y = -\frac{x^3}{3} + \frac{5x^2}{2} - 4x + 8\).

To find the equation of the curve, we need to integrate the derivative \(\frac{dy}{dx} = \frac{8}{\sqrt{4x + 1}}\).

The integral of \(\frac{8}{\sqrt{4x + 1}}\) is:

\(y = 8 \int (4x + 1)^{-\frac{1}{2}} \, dx\)

Using the substitution \(u = 4x + 1\), \(du = 4 \, dx\), so \(dx = \frac{1}{4} \, du\).

Thus, the integral becomes:

\(y = 8 \int u^{-\frac{1}{2}} \cdot \frac{1}{4} \, du\)

\(y = 2 \int u^{-\frac{1}{2}} \, du\)

\(y = 2 \cdot 2u^{\frac{1}{2}} + c\)

\(y = 4\sqrt{4x + 1} + c\)

Using the point \((2, 5)\) to find \(c\):

\(5 = 4\sqrt{4(2) + 1} + c\)

\(5 = 4\sqrt{9} + c\)

\(5 = 12 + c\)

\(c = -7\)

Therefore, the equation of the curve is \(y = 4\sqrt{4x+1} - 7\).

To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = \frac{8}{(5 - 2x)^2}\).

Let \(f(x) = \int \frac{8}{(5 - 2x)^2} \, dx\).

Using substitution, let \(u = 5 - 2x\), then \(\frac{du}{dx} = -2\) or \(dx = -\frac{1}{2} \, du\).

Substitute into the integral:

\(f(x) = \int \frac{8}{u^2} \left(-\frac{1}{2}\right) \, du = -4 \int u^{-2} \, du\).

Integrate: \(f(x) = -4 \left( \frac{u^{-1}}{-1} \right) + c = \frac{4}{u} + c\).

Substitute back \(u = 5 - 2x\):

\(f(x) = \frac{4}{5 - 2x} + c\).

Given the curve passes through (2, 7), substitute \(x = 2\) and \(y = 7\):

\(7 = \frac{4}{5 - 2(2)} + c\).

\(7 = \frac{4}{1} + c\).

\(7 = 4 + c\).

\(c = 3\).

Thus, the equation of the curve is \(y = \frac{4}{5 - 2x} + 3\).

To find the equation of the curve, we need to integrate the derivative \(\frac{dy}{dx} = 3x^2 - \frac{2}{x^3}\).

Integrating term by term, we have:

\(y = \int (3x^2) \, dx - \int \left(\frac{2}{x^3}\right) \, dx\)

\(y = \frac{3x^3}{3} - \frac{2x^{-2}}{-2} + c\)

\(y = x^3 + x^{-2} + c\)

Since the curve passes through \((-1, 3)\), substitute \(x = -1\) and \(y = 3\) to find \(c\):

\(3 = (-1)^3 + (-1)^{-2} + c\)

\(3 = -1 + 1 + c\)

\(3 = c\)

Thus, the equation of the curve is \(y = x^3 + x^{-2} + 3\).

(a) Given \(\frac{d^2y}{dx^2} = 6x\), integrate to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \int 6x \, dx = 3x^2 + c\).

Since the curve has a stationary point at \((2, -10)\), \(\frac{dy}{dx} = 0\) when \(x = 2\):

\(3(2)^2 + c = 0\)

\(12 + c = 0\)

\(c = -12\)

Thus, \(\frac{dy}{dx} = 3x^2 - 12\).

(b) Integrate \(\frac{dy}{dx} = 3x^2 - 12\) to find \(y\):

\(y = \int (3x^2 - 12) \, dx = x^3 - 12x + k\).

Substitute \((2, -10)\) into the equation:

\(-10 = 2^3 - 12 \times 2 + k\)

\(-10 = 8 - 24 + k\)

\(-10 = -16 + k\)

\(k = 6\)

Thus, \(y = x^3 - 12x + 6\).

(c) Set \(\frac{dy}{dx} = 0\) to find other stationary points:

\(3x^2 - 12 = 0\)

\(3x^2 = 12\)

\(x^2 = 4\)

\(x = \pm 2\)

Since \(x = 2\) is already given, consider \(x = -2\):

\(y = (-2)^3 - 12(-2) + 6 = -8 + 24 + 6 = 22\)

So, the point is \((-2, 22)\).

Check the nature using \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = 6x\)

At \(x = -2\), \(\frac{d^2y}{dx^2} = 6(-2) = -12 < 0\), hence a maximum.

(d) Find the equation of the tangent at the y-intercept \((0, y)\):

\(y = 0^3 - 12 \times 0 + 6 = 6\)

The slope \(\frac{dy}{dx}\) at \(x = 0\) is:

\(\frac{dy}{dx} = 3(0)^2 - 12 = -12\)

Equation of the tangent: \(y - 6 = -12(x - 0)\)

\(y = -12x + 6\)

To find \(f(x)\), we need to integrate \(f'(x) = 3x^2 - 7\).

Integrating, we get:

\(f(x) = \int (3x^2 - 7) \, dx = x^3 - 7x + c\)

We know \(f(3) = 5\), so substitute \(x = 3\) into the equation:

\(5 = 3^3 - 7 \times 3 + c\)

\(5 = 27 - 21 + c\)

\(5 = 6 + c\)

\(c = -1\)

Thus, \(f(x) = x^3 - 7x - 1\).

To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = (2x + 1)^{\frac{1}{2}}\).

Integrate: \(y = \int (2x + 1)^{\frac{1}{2}} \, dx\).

Let \(u = 2x + 1\), then \(du = 2 \, dx\) or \(dx = \frac{1}{2} \, du\).

Substitute: \(y = \int u^{\frac{1}{2}} \cdot \frac{1}{2} \, du = \frac{1}{2} \int u^{\frac{1}{2}} \, du\).

Integrate: \(\frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}} = \frac{1}{3} (2x + 1)^{\frac{3}{2}} + c\).

Thus, \(y = \frac{(2x + 1)^{\frac{3}{2}}}{3} + c\).

Use the point \((4, 7)\) to find \(c\):

\(7 = \frac{(2 \cdot 4 + 1)^{\frac{3}{2}}}{3} + c\).

\(7 = \frac{9^{\frac{3}{2}}}{3} + c\).

\(7 = \frac{27}{3} + c\).

\(7 = 9 + c\).

\(c = -2\).

Therefore, the equation of the curve is \(y = \frac{(2x + 1)^{\frac{3}{2}}}{3} - 2\).

To find \(f(x)\), we need to integrate \(f'(x) = 5 - 2x^2\).

\(\int (5 - 2x^2) \, dx = \int 5 \, dx - \int 2x^2 \, dx\)

\(= 5x - \frac{2x^3}{3} + c\)

We know that \((3, 5)\) is a point on the curve, so \(f(3) = 5\).

Substitute \(x = 3\) and \(f(x) = 5\) into the equation:

\(5 = 5(3) - \frac{2(3)^3}{3} + c\)

\(5 = 15 - \frac{54}{3} + c\)

\(5 = 15 - 18 + c\)

\(5 = -3 + c\)

\(c = 8\)

Thus, \(f(x) = 5x - \frac{2x^3}{3} + 8\).

To find \(f(x)\), we need to integrate \(f'(x) = x^{-\frac{3}{2}} + 1\).

The integral of \(x^{-\frac{3}{2}}\) is \(2x^{-\frac{1}{2}}\), and the integral of \(1\) is \(x\).

Thus, \(f(x) = 2x^{\frac{1}{2}} + x + c\), where \(c\) is a constant.

We use the condition \(f(4) = 5\) to find \(c\):

\(5 = 2 \times 4^{\frac{1}{2}} + 4 + c\)

\(5 = 2 \times 2 + 4 + c\)

\(5 = 4 + 4 + c\)

\(5 = 8 + c\)

\(c = 5 - 8 = -3\)

Therefore, \(f(x) = 2x^{\frac{1}{2}} + x - 3\).

To find \(f(x)\), we need to integrate \(f'(x) = \frac{1}{\sqrt{x+6}} + \frac{6}{x^2}\).

The integral of \(\frac{1}{\sqrt{x+6}}\) is \(2\sqrt{x+6}\).

The integral of \(\frac{6}{x^2}\) is \(-\frac{6}{x}\).

Thus, \(f(x) = 2\sqrt{x+6} - \frac{6}{x} + c\).

Using the condition \(f(3) = 1\), substitute \(x = 3\) into the equation:

\(2\sqrt{3+6} - \frac{6}{3} + c = 1\).

This simplifies to \(2(3) - 2 + c = 1\).

\(6 - 2 + c = 1\).

\(c = -3\).

Therefore, \(f(x) = 2\sqrt{x+6} - \frac{6}{x} - 3\).

Given \(\frac{dy}{dx} = \sqrt{2x + 5}\), we need to find \(y\) by integrating.

Integrate \(\frac{dy}{dx} = \sqrt{2x + 5} = (2x + 5)^{1/2}\).

\(y = \int (2x + 5)^{1/2} \, dx\)

Using the substitution \(u = 2x + 5\), \(du = 2 \, dx\), so \(dx = \frac{1}{2} \, du\).

\(y = \int (u)^{1/2} \cdot \frac{1}{2} \, du\)

\(y = \frac{1}{2} \int u^{1/2} \, du\)

\(y = \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + c\)

\(y = \frac{1}{3} (2x + 5)^{3/2} + c\)

Given the point \((2, 5)\) is on the curve, substitute \(x = 2\) and \(y = 5\):

\(5 = \frac{1}{3} (2(2) + 5)^{3/2} + c\)

\(5 = \frac{1}{3} (4 + 5)^{3/2} + c\)

\(5 = \frac{1}{3} (9)^{3/2} + c\)

\(5 = \frac{1}{3} \cdot 27 + c\)

\(5 = 9 + c\)

\(c = -4\)

Thus, the equation of the curve is:

\(y = \frac{1}{3} (2x + 5)^{3/2} - 4\)

\(y = \frac{2}{3} (2x + 5)^{3/2} - 4\)

Given \(\frac{dy}{dx} = \frac{6}{x^2}\), we need to find \(y\) by integrating \(\frac{dy}{dx}\).

Integrate \(\frac{dy}{dx} = \frac{6}{x^2}\):

\(y = \int \frac{6}{x^2} \, dx = \int 6x^{-2} \, dx\)

\(y = 6 \cdot \left( \frac{x^{-1}}{-1} \right) + c = -6x^{-1} + c\)

Using the point \((2, 9)\) to find \(c\):

\(9 = -6(2)^{-1} + c\)

\(9 = -3 + c\)

\(c = 12\)

Thus, the equation of the curve is \(y = -6x^{-1} + 12\).

To find \(f(x)\), we need to integrate \(f'(x) = 3x^{\frac{1}{2}} + 3x^{-\frac{1}{2}} - 10\).

Integrating term by term:

\(\int 3x^{\frac{1}{2}} \, dx = 2x^{\frac{3}{2}} + C_1\)

\(\int 3x^{-\frac{1}{2}} \, dx = 6x^{\frac{1}{2}} + C_2\)

\(\int -10 \, dx = -10x + C_3\)

Thus, \(f(x) = 2x^{\frac{3}{2}} + 6x^{\frac{1}{2}} - 10x + c\).

We know the curve passes through the point \((4, -7)\), so substitute \(x = 4\) and \(f(x) = -7\):

\(-7 = 2(4)^{\frac{3}{2}} + 6(4)^{\frac{1}{2}} - 10(4) + c\)

\(-7 = 16 + 12 - 40 + c\)

\(-7 = -12 + c\)

\(c = 5\)

Therefore, \(f(x) = 2x^{\frac{3}{2}} + 6x^{\frac{1}{2}} - 10x + 5\).

To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = 2(3x + 4)^{\frac{3}{2}} - 6x - 8\).

Integrate each term separately:

1. \(\int 2(3x + 4)^{\frac{3}{2}} \, dx\)

Let \(u = 3x + 4\), then \(du = 3 \, dx\) or \(dx = \frac{du}{3}\).

\(\int 2(3x + 4)^{\frac{3}{2}} \, dx = \frac{2}{3} \int u^{\frac{3}{2}} \, du = \frac{2}{3} \cdot \frac{2}{5} u^{\frac{5}{2}} = \frac{4}{15} (3x + 4)^{\frac{5}{2}}\)

2. \(\int -6x \, dx = -3x^2\)

3. \(\int -8 \, dx = -8x\)

Thus, the integral is:

\(y = \frac{4}{15} (3x + 4)^{\frac{5}{2}} - 3x^2 - 8x + c\)

Given the stationary point \((-1, 5)\), substitute into the equation:

\(5 = \frac{4}{15}(3(-1) + 4)^{\frac{5}{2}} - 3(-1)^2 - 8(-1) + c\)

\(5 = \frac{4}{15}(1)^{\frac{5}{2}} - 3 + 8 + c\)

\(5 = \frac{4}{15} + 5 + c\)

\(c = 5 - 5 - \frac{4}{15} = -\frac{4}{15}\)

Therefore, the equation of the curve is:

\(y = \frac{4}{15}(3x + 4)^{\frac{5}{2}} - 3x^2 - 8x - \frac{4}{15}\)

To find the equation of the curve, we need to integrate the given derivative \(\frac{dy}{dx} = -\frac{8}{x^3} - 1\).

Integrate \(\frac{dy}{dx}\) with respect to \(x\):

\(y = \int \left( -\frac{8}{x^3} - 1 \right) \, dx\)

\(y = \int -8x^{-3} \, dx - \int 1 \, dx\)

\(y = -8 \int x^{-3} \, dx - \int 1 \, dx\)

\(y = -8 \left( \frac{x^{-2}}{-2} \right) - x + c\)

\(y = 4x^{-2} - x + c\)

\(y = \frac{4}{x^2} - x + c\)

Given that the point (2, 4) lies on the curve, substitute \(x = 2\) and \(y = 4\) to find \(c\):

\(4 = \frac{4}{2^2} - 2 + c\)

\(4 = 1 - 2 + c\)

\(4 = -1 + c\)

\(c = 5\)

Thus, the equation of the curve is \(y = \frac{4}{x^2} - x + 5\).

(a) To find the equation of the curve, integrate \(f'(x) = 1 - \frac{2}{(x-1)^3}\).

\(\int \left(1 - \frac{2}{(x-1)^3}\right) \, dx = x + \int -\frac{2}{(x-1)^3} \, dx\)

Let \(u = x-1\), then \(du = dx\). The integral becomes \(\int -\frac{2}{u^3} \, du = \frac{1}{u^2} = \frac{1}{(x-1)^2}\).

Thus, \(y = x + \frac{1}{(x-1)^2} + c\).

Given \((0, 3)\), substitute to find \(c\):

\(3 = 0 + \frac{1}{1} + c \Rightarrow c = 2\).

Therefore, \(y = x + (x-1)^2 + 2\).

(b) The gradient of the tangent at \(x = 0\) is \(f'(0) = 3\).

The equation of the tangent is \(y - 3 = 3(x - 0) \Rightarrow y = 3x + 3\).

Set \(3x + 3 = x + (x-1)^2 + 2\) to find intersection:

\(3x + 3 = x + (x-1)^2 + 2 \Rightarrow 2x + 1 = \frac{1}{(x-1)^2}\).

Multiply through by \((x-1)^2\):

\((2x + 1)(x-1)^2 = 1 \Rightarrow (2x + 1)(x-1)^2 - 1 = 0\).

(c) Substitute \(x = \frac{3}{2}\) into \((2x + 1)(x-1)^2 - 1 = 0\):

\((2 \times \frac{3}{2} + 1)(\frac{3}{2} - 1)^2 - 1 = 0\).

\((3 + 1)(\frac{1}{2})^2 - 1 = 0 \Rightarrow 4 \times \frac{1}{4} - 1 = 0\).

Thus, \(x = \frac{3}{2}\) satisfies the equation.

Find \(y\)-coordinate: \(y = \frac{3}{2} + (\frac{3}{2} - 1)^2 + 2\).

\(y = \frac{3}{2} + \frac{1}{4} + 2 = \frac{7}{2}\).

Given \(\frac{dy}{dx} = \frac{2}{\sqrt{x}} - 1\), we need to find \(y\) by integrating.

Integrate \(\frac{dy}{dx} = \frac{2}{\sqrt{x}} - 1\):

\(y = \int \left( \frac{2}{\sqrt{x}} - 1 \right) \, dx\)

\(y = \int 2x^{-1/2} \, dx - \int 1 \, dx\)

\(y = 2 \cdot \frac{x^{1/2}}{1/2} - x + c\)

\(y = 4\sqrt{x} - x + c\)

Use the point \(P(9, 5)\) to find \(c\):

\(5 = 4\sqrt{9} - 9 + c\)

\(5 = 12 - 9 + c\)

\(5 = 3 + c\)

\(c = 2\)

Thus, the equation of the curve is \(y = 4\sqrt{x} - x + 2\).

To find \(f(x)\), we need to integrate \(f'(x) = 3x^2 + 2x - 5\).

The integral of \(3x^2\) is \(x^3\), the integral of \(2x\) is \(x^2\), and the integral of \(-5\) is \(-5x\). Therefore,

\(f(x) = x^3 + x^2 - 5x + c\), where \(c\) is the constant of integration.

We use the point \((1, 3)\) to find \(c\). Substituting \(x = 1\) and \(f(x) = 3\) into the equation:

\(3 = 1^3 + 1^2 - 5(1) + c\)

\(3 = 1 + 1 - 5 + c\)

\(3 = -3 + c\)

\(c = 6\)

Thus, \(f(x) = x^3 + x^2 - 5x + 6\).

To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = \frac{3}{\sqrt{x}} - x\).

First, integrate \(\frac{3}{\sqrt{x}}\):

\(\int \frac{3}{\sqrt{x}} \, dx = \int 3x^{-1/2} \, dx = 6\sqrt{x} + C_1\).

Next, integrate \(-x\):

\(\int -x \, dx = -\frac{x^2}{2} + C_2\).

Combining these, the general solution is:

\(y = 6\sqrt{x} - \frac{x^2}{2} + C\).

Given the point (4, 6), substitute \(x = 4\) and \(y = 6\) to find \(C\):

\(6 = 6\sqrt{4} - \frac{4^2}{2} + C\).

\(6 = 12 - 8 + C\).

\(C = 2\).

Thus, the equation of the curve is \(y = 6\sqrt{x} - \frac{x^2}{2} + 2\).

To find the equation of the curve, we need to integrate the derivative \(\frac{dy}{dx} = 2x^2 - 5\).

Integrating, we get:

\(y = \int (2x^2 - 5) \, dx = \frac{2x^3}{3} - 5x + c\)

We use the point \((3, 8)\) to find the constant \(c\).

Substitute \(x = 3\) and \(y = 8\) into the equation:

\(8 = \frac{2(3)^3}{3} - 5(3) + c\)

\(8 = \frac{54}{3} - 15 + c\)

\(8 = 18 - 15 + c\)

\(8 = 3 + c\)

\(c = 5\)

Thus, the equation of the curve is:

\(y = \frac{2x^3}{3} - 5x + 5\)

To find the equation of the curve, we need to integrate \(\frac{dy}{dx} = \frac{4}{(x-3)^3}\).

Integrate: \(y = \int \frac{4}{(x-3)^3} \, dx\).

Let \(u = x-3\), then \(du = dx\).

\(y = \int \frac{4}{u^3} \, du = 4 \int u^{-3} \, du\).

Integrating, we get \(y = 4 \left( \frac{u^{-2}}{-2} \right) + c = \frac{-2}{u^2} + c\).

Substitute back \(u = x-3\): \(y = \frac{-2}{(x-3)^2} + c\).

Use the point (4, 5) to find \(c\):

\(5 = \frac{-2}{(4-3)^2} + c\).

\(5 = -2 + c\).

\(c = 7\).

Thus, the equation of the curve is \(y = \frac{-2}{(x-3)^2} + 7\).

(a) To find the value of \(a\), set \(\frac{dy}{dx} = 0\) at the stationary point \(x = a\). Thus, \(6a^2 - 30a + 6a = 0\). Simplifying gives \(6a(a - 4) = 0\), so \(a = 4\).

(b) Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2} = 12x - 30\). At \(x = 4\), \(\frac{d^2y}{dx^2} = 18\), which is positive, indicating a minimum.

(c) Integrate \(\frac{dy}{dx} = 6x^2 - 30x + 24\) to find \(y\). The integral is \(y = 2x^3 - 15x^2 + 24x + c\). Using the point \((4, -15)\), solve for \(c\): \(-15 = 2(4)^3 - 15(4)^2 + 24(4) + c\), giving \(c = 1\). Thus, \(y = 2x^3 - 15x^2 + 24x + 1\).

(d) Set \(\frac{dy}{dx} = 0\) to find other stationary points: \(6x^2 - 30x + 24 = 0\). Factor to get \(6(x-1)(x-4) = 0\), so \(x = 1\) or \(x = 4\). For \(x = 1\), \(y = 2(1)^3 - 15(1)^2 + 24(1) + 1 = 12\). Thus, the coordinates are \((1, 12)\).

(a) Given \(\frac{dy}{dx} = x^{-\frac{1}{2}} + k\) and the gradient at (4, -1) is \(-\frac{3}{2}\), substitute \(x = 4\):

\(-\frac{3}{2} = 4^{-\frac{1}{2}} + k\)

\(-\frac{3}{2} = \frac{1}{2} + k\)

Solving for \(k\), we get \(k = -2\).

(b) Integrate \(\frac{dy}{dx} = x^{-\frac{1}{2}} - 2\):

\(y = \int (x^{-\frac{1}{2}} - 2) \, dx = 2x^{\frac{1}{2}} - 2x + c\)

Substitute \(x = 4, y = -1\):

\(-1 = 2(4)^{\frac{1}{2}} - 2(4) + c\)

\(-1 = 4 - 8 + c\)

\(c = 3\)

Thus, the equation is \(y = 2x^{\frac{1}{2}} - 2x + 3\).

(c) Set \(\frac{dy}{dx} = 0\):

\(x^{-\frac{1}{2}} - 2 = 0\)

\(x^{-\frac{1}{2}} = 2\)

\(x = \frac{1}{4}\)

Substitute \(x = \frac{1}{4}\) into the equation:

\(y = 2\left(\frac{1}{4}\right)^{\frac{1}{2}} - 2\left(\frac{1}{4}\right) + 3\)

\(y = 1 - \frac{1}{2} + 3 = \frac{3}{2}\)

The stationary point is \(\left(\frac{1}{4}, \frac{3}{2}\right)\).

(d) Find \(\frac{d^2y}{dx^2}\):

\(\frac{d^2y}{dx^2} = -\frac{1}{2}x^{-\frac{3}{2}}\)

At \(x = \frac{1}{4}\), \(\frac{d^2y}{dx^2} = -\frac{1}{2}\left(\frac{1}{4}\right)^{-\frac{3}{2}} = -4\)

Since \(\frac{d^2y}{dx^2} < 0\), the stationary point is a maximum.

(a) We have \(f'(x) = \frac{-3}{(x+2)^4}\) and the gradient of the tangent is \(-\frac{16}{27}\) at \(x = a\). Set \(\frac{-3}{(a+2)^4} = -\frac{16}{27}\).

Equating gives \(16(a+2)^4 = 81\).

Solving for \((a+2)^2\), we get \((a+2)^2 = \frac{9}{4}\).

Thus, \(a+2 = \pm \frac{3}{2}\).

Solving for \(a\), we find \(a = \frac{1}{2}\) or \(a = -\frac{7}{2}\).

(b) To find \(f(x)\), integrate \(f'(x) = \frac{-3}{(x+2)^4}\).

\(f(x) = \int \frac{-3}{(x+2)^4} \, dx = \frac{-1}{(x+2)^3} + c\).

Given \(f(-1) = 5\), substitute to find \(c\):

\(5 = \frac{-1}{1^3} + c \Rightarrow 5 = -1 + c \Rightarrow c = 6\).

Thus, \(f(x) = \frac{-1}{(x+2)^3} + 6\).

(a) To find the equation of the curve, integrate \(\frac{dy}{dx} = 3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}}\).

Integrating, we get:

\(y = \int (3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}}) \, dx\)

\(y = \left[ \frac{3}{\frac{3}{2}} x^{\frac{3}{2}} \right] - \left[ \frac{3}{\frac{1}{2}} x^{\frac{1}{2}} \right] + c\)

\(y = 2x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + c\)

Using the point \((3, 5)\) to find \(c\):

\(5 = 2(3)^{\frac{3}{2}} - 6(3)^{\frac{1}{2}} + c\)

\(5 = 2 \times 3\sqrt{3} - 6 \times \sqrt{3} + c\)

\(5 = 6\sqrt{3} - 6\sqrt{3} + c\)

\(c = 5\)

Thus, the equation is \(y = 2x^{\frac{3}{2}} - 6x^{\frac{1}{2}} + 5\).

(b) To find the \(x\)-coordinate of the stationary point, set \(\frac{dy}{dx} = 0\):

\(3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}} = 0\)

\(3x^{\frac{1}{2}} = 3x^{-\frac{1}{2}}\)

\(x^{\frac{1}{2}} = x^{-\frac{1}{2}}\)

\(x = 1\)

(c) \(y\) increases as \(x\) increases when \(\frac{dy}{dx} > 0\):

\(3x^{\frac{1}{2}} - 3x^{-\frac{1}{2}} > 0\)

\(x^{\frac{1}{2}} > x^{-\frac{1}{2}}\)

\(x > 1\)