Answer: stationary point \((-2,0)\), minimum.

We are given

\(\mathrm f''(x)=(2x+5)^{-\frac32}.\)

Integrate once to find \(\mathrm f'(x)\):

\(\mathrm f'(x)=\int(2x+5)^{-\frac32}\,\mathrm dx.\)

Since \(\dfrac{\mathrm d}{\mathrm dx}(2x+5)=2\),

\(\int(2x+5)^{-\frac32}\,\mathrm dx=-(2x+5)^{-\frac12}+c.\)

So

\(\mathrm f'(x)=1-\dfrac1{\sqrt{2x+5}}\)

after finding the constant as follows. The gradient is \(\dfrac23\) when \(x=2\), so

\(-9^{-\frac12}+c=\dfrac23.\)

Thus

\(-\dfrac13+c=\dfrac23,\)

giving \(c=1\).

At a stationary point, \(\mathrm f'(x)=0\), so

\(1-\dfrac1{\sqrt{2x+5}}=0.\)

Hence

\(\sqrt{2x+5}=1,\)

so

\(2x+5=1,\qquad x=-2.\)

Now integrate \(\mathrm f'(x)\) to find \(\mathrm f(x)\):

\(\mathrm f(x)=\int\left(1-(2x+5)^{-\frac12}\right)\,\mathrm dx.\)

This gives

\(\mathrm f(x)=x-\sqrt{2x+5}+d.\)

The curve passes through \((2,2)\), so

\(2=2-\sqrt9+d.\)

Thus \(d=3\), and

\(\mathrm f(x)=x-\sqrt{2x+5}+3.\)

At \(x=-2\),

\(\mathrm f(-2)=-2-1+3=0.\)

Therefore the stationary point is

\((-2,0).\)

To determine the nature, use the second derivative:

\(\mathrm f''(-2)=(2(-2)+5)^{-\frac32}=1\gt0.\)

So the stationary point is a minimum.

Answer: \(\displaystyle f(x)=\frac1{24}(5x+2)^{8/5}+3x-20\).

Differentiate first. Stationary points occur where the derivative is zero, while tangent and normal problems use the gradient at the given point.

We are given

\(f''(x)=(5x+2)^{-2/5}.\)

Integrating once,

\(f'(x)=\frac13(5x+2)^{3/5}+c.\)

Use \(f'(6)=\frac{17}{3}\). Since \(5(6)+2=32\),

\(\frac{17}{3}=\frac13(32)^{3/5}+c.\)

Now \(32^{3/5}=8\), so

\(\frac{17}{3}=\frac83+c.\)

Hence

\(c=3.\)

Therefore

\(f'(x)=\frac13(5x+2)^{3/5}+3.\)

Integrating again,

\(f(x)=\frac1{24}(5x+2)^{8/5}+3x+d.\)

Use \(f(6)=\frac{26}{3}\). Again \(5(6)+2=32\), and \(32^{8/5}=256\), so

\(\frac{26}{3}=\frac1{24}(256)+18+d.\)

Thus

\(\frac{26}{3}=\frac{32}{3}+18+d.\)

Since \(18=\frac{54}{3}\),

\(\frac{26}{3}=\frac{86}{3}+d.\)

So

\(d=-20.\)

Therefore

\(f(x)=\frac1{24}(5x+2)^{8/5}+3x-20.\)

Answer: \(\displaystyle f(x)=\frac89(3x+4)^{3/2}+2x-8\).

Answer: \(\displaystyle f(x)=\frac89(3x+4)^{3/2}+2x-8\).

We are given

\(f''(x)=6(3x+4)^{-1/2}.\)

Integrate to find \(f'(x)\):

\(f'(x)=4(3x+4)^{1/2}+C.\)

Use \(f'(4)=18\):

\(18=4(3(4)+4)^{1/2}+C.\)

Since \(3(4)+4=16\),

\(18=4\sqrt{16}+C=16+C.\)

Thus

\(C=2.\)

So

\(f'(x)=4(3x+4)^{1/2}+2.\)

Integrate again:

\(f(x)=\frac89(3x+4)^{3/2}+2x+D.\)

Use \(f(4)=\frac{512}{9}\):

\(\frac{512}{9}=\frac89(16)^{3/2}+2(4)+D.\)

Now \(16^{3/2}=64\), so

\(\frac{512}{9}=\frac89(64)+8+D.\)

Since \(\frac89(64)=\frac{512}{9}\),

\(D=-8.\)

Therefore

\(f(x)=\frac89(3x+4)^{3/2}+2x-8.\)

Answer: \(\frac{dy}{dx}=(2x+3)^{1/2}+5x-8\); \(\mathrm f(x)=\frac13(2x+3)^{3/2}+\frac52x^2-8x+2\).

(a) Since

\(\frac{d^2y}{dx^2}=(2x+3)^{-\frac12}+5,\)

integrate once to find the gradient:

\(\frac{dy}{dx}=(2x+3)^{\frac12}+5x+C.\)

The curve has gradient \(10\) when \(x=3\), so

\(10=(2(3)+3)^{\frac12}+5(3)+C.\)

Thus

\(10=3+15+C,\)

so \(C=-8\). Therefore

\(\frac{dy}{dx}=(2x+3)^{\frac12}+5x-8.\)

When \(x=11\),

\(\frac{dy}{dx}=(25)^{\frac12}+55-8=5+55-8=52.\)

(b) Integrate the expression for the gradient:

Since

we get

The curve passes through \(\left(3,\frac{19}{2}\right)\), so

This gives

\(\frac{19}{2}=9+\frac{45}{2}-24+D.\)

So

\(D=2.\)

Therefore

Answer: \(\frac{dy}{dx}=\frac12e^{2x}-\frac1{x+1}+\frac52\); \(y=\frac14e^{2x}-\ln(x+1)+\frac52x+\frac{15}{4}\).

(a) Integrate

\(\frac{d^2y}{dx^2}=e^{2x}+\frac1{(x+1)^2}.\)

This gives

\(\frac{dy}{dx}=\frac12e^{2x}-(x+1)^{-1}+C.\)

Since \(\frac{dy}{dx}=2\) when \(x=0\),

\(2=\frac12e^0-1+C.\)

So

\(2=\frac12-1+C,\)

which gives

\(C=\frac52.\)

Therefore

\(\frac{dy}{dx}=\frac12e^{2x}-\frac1{x+1}+\frac52.\)

(b) Integrate again:

\(y=\frac14e^{2x}-\ln(x+1)+\frac52x+D.\)

Since \(y=4\) when \(x=0\),

\(4=\frac14-\ln1+0+D.\)

Thus

\(D=\frac{15}{4}.\)

Hence

\(y=\frac14e^{2x}-\ln(x+1)+\frac52x+\frac{15}{4}.\)

Answer: \(y=-\frac54\cos2x+2x+\frac{17\pi}{12}+\frac{5\sqrt3}{8}\).

Given that

\(\frac{d^2y}{dx^2}=5\cos2x,\)

integrate once to find the gradient:

\(\frac{dy}{dx}=\frac52\sin2x+c.\)

The gradient is \(\frac34\) when \(x=-\frac{\pi}{12}\), so

\(\frac34=\frac52\sin\left(-\frac{\pi}{6}\right)+c.\)

Since \(\sin\left(-\frac{\pi}{6}\right)=-\frac12\),

\(\frac34=-\frac54+c.\)

Hence

\(c=2.\)

So

\(\frac{dy}{dx}=\frac52\sin2x+2.\)

Integrate again:

\(y=-\frac54\cos2x+2x+d.\)

The curve passes through

\(\left(-\frac{\pi}{12},\frac{5\pi}{4}\right).\)

Substitute these values:

\(\frac{5\pi}{4} =-\frac54\cos\left(-\frac{\pi}{6}\right)+2\left(-\frac{\pi}{12}\right)+d.\)

Since \(\cos\left(-\frac{\pi}{6}\right)=\frac{\sqrt3}{2}\),

\(\frac{5\pi}{4} =-\frac{5\sqrt3}{8}-\frac{\pi}{6}+d.\)

Therefore

\(d=\frac{5\pi}{4}+\frac{\pi}{6}+\frac{5\sqrt3}{8}.\)

So

\(d=\frac{17\pi}{12}+\frac{5\sqrt3}{8}.\)

The equation of the curve is

\(y=-\frac54\cos2x+2x+\frac{17\pi}{12}+\frac{5\sqrt3}{8}.\)

Answer: \(y=-2\sin\left(x+\frac{\pi}{3}\right)+4x+\frac{\pi}{3}+\sqrt3\).

Integrate once:

\(\frac{dy}{dx}=\int 2\sin\left(x+\frac{\pi}{3}\right)\,dx.\)

Therefore

\(\frac{dy}{dx}=-2\cos\left(x+\frac{\pi}{3}\right)+c.\)

The gradient is \(5\) when \(x=\frac{\pi}{3}\), so

\(5=-2\cos\left(\frac{2\pi}{3}\right)+c.\)

Since \(\cos\frac{2\pi}{3}=-\frac12\),

\(5=1+c,\)

and hence

\(c=4.\)

So

\(\frac{dy}{dx}=-2\cos\left(x+\frac{\pi}{3}\right)+4.\)

Integrate again:

\(y=-2\sin\left(x+\frac{\pi}{3}\right)+4x+d.\)

The curve passes through \(\left(\frac{\pi}{3},\frac{5\pi}{3}\right)\), so

\(\frac{5\pi}{3}=-2\sin\left(\frac{2\pi}{3}\right)+\frac{4\pi}{3}+d.\)

Since \(\sin\frac{2\pi}{3}=\frac{\sqrt3}{2}\), this gives

\(\frac{5\pi}{3}=-\sqrt3+\frac{4\pi}{3}+d.\)

Thus

\(d=\frac{\pi}{3}+\sqrt3.\)

Therefore the equation of the curve is

\(\boxed{y=-2\sin\left(x+\frac{\pi}{3}\right)+4x+\frac{\pi}{3}+\sqrt3}.\)

Answer: \(y=\frac12(3x-1)^{\frac43}+2x-3\).

Integrate once:

\(\frac{dy}{dx}=\int 2(3x-1)^{-\frac23}\,dx.\)

Let \(u=3x-1\), so \(du=3\,dx\). Then

\(\frac{dy}{dx}=2(3x-1)^{\frac13}+c.\)

The gradient is \(6\) when \(x=3\). Since \(3x-1=8\),

\(6=2(8)^{\frac13}+c.\)

So

\(6=4+c,\)

and hence

\(c=2.\)

Therefore

\(\frac{dy}{dx}=2(3x-1)^{\frac13}+2.\)

Integrate again:

\(y=\frac12(3x-1)^{\frac43}+2x+d.\)

The curve passes through \((3,11)\), so

\(11=\frac12(8)^{\frac43}+6+d.\)

Now \(8^{1/3}=2\), so \(8^{4/3}=16\). Hence

\(11=8+6+d,\)

giving

\(d=-3.\)

So the equation of the curve is

\(\boxed{y=\frac12(3x-1)^{\frac43}+2x-3}.\)

Answer: \(y=\dfrac13(2x-5)^{3/2}+4x-20\).

Integrate the second derivative once:

\(\frac{dy}{dx} = \int (2x-5)^{-1/2}\,dx.\)

Since the derivative of \(2x-5\) is \(2\),

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

So

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

The gradient is \(6\) when \(x=\dfrac92\). Substitute this:

\(6=\left(2\cdot\frac92-5\right)^{1/2}+c.\)

Now

\(2\cdot\frac92-5=4,\)

so

\(6=2+c.\)

Hence \(c=4\), and

\(\frac{dy}{dx}=(2x-5)^{1/2}+4.\)

Integrate again:

\(y=\int\left((2x-5)^{1/2}+4\right)\,dx.\)

This gives

\(y=\frac13(2x-5)^{3/2}+4x+d.\)

Use the point \(\left(\dfrac92,\dfrac23\right)\):

\(\frac23 = \frac13(4)^{3/2}+4\left(\frac92\right)+d.\)

Since \(4^{3/2}=8\),

\(\frac23=\frac83+18+d.\)

Therefore

\(d=-20.\)

Hence the equation of the curve is

\(\boxed{y=\frac13(2x-5)^{3/2}+4x-20}.\)

Answer: \(y=\dfrac13(2x-5)^{3/2}+4x-20\).

Integrate the second derivative once:

\(\frac{dy}{dx} = \int (2x-5)^{-1/2}\,dx.\)

Since the derivative of \(2x-5\) is \(2\),

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

So

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

The gradient is \(6\) when \(x=\dfrac92\). Substitute this:

\(6=\left(2\cdot\frac92-5\right)^{1/2}+c.\)

Now

\(2\cdot\frac92-5=4,\)

so

\(6=2+c.\)

Hence \(c=4\), and

\(\frac{dy}{dx}=(2x-5)^{1/2}+4.\)

Integrate again:

\(y=\int\left((2x-5)^{1/2}+4\right)\,dx.\)

This gives

\(y=\frac13(2x-5)^{3/2}+4x+d.\)

Use the point \(\left(\dfrac92,\dfrac23\right)\):

\(\frac23 = \frac13(4)^{3/2}+4\left(\frac92\right)+d.\)

Since \(4^{3/2}=8\),

\(\frac23=\frac83+18+d.\)

Therefore

\(d=-20.\)

Hence the equation of the curve is

\(\boxed{y=\frac13(2x-5)^{3/2}+4x-20}.\)

Answer: \(y=\dfrac52x^2+\dfrac14e^{2x}+\dfrac72x-\dfrac{13}{4}\).

Integrate once:

\(\frac{dy}{dx}=5x+\frac12e^{2x}+C.\)

At \(x=0\), the gradient is \(4\), so

\(4=\frac12+C.\)

Hence \(C=\frac72\), and

\(\frac{dy}{dx}=5x+\frac12e^{2x}+\frac72.\)

Integrate again:

\(y=\frac52x^2+\frac14e^{2x}+\frac72x+D.\)

The curve passes through \((0,-3)\), so

\(-3=\frac14+D.\)

Therefore \(D=-\frac{13}{4}\), and the equation of the curve is

\(\boxed{y=\frac52x^2+\frac14e^{2x}+\frac72x-\frac{13}{4}}.\)

Answer: (i) \(\dfrac{dy}{dx}=x^2-\dfrac{1}{(x+1)^3}+\dfrac18\). (ii) \(y=\dfrac{x^3}{3}+\dfrac{1}{2(x+1)^2}+\dfrac{x}{8}+\dfrac{29}{12}\).

(i) Integrate once:

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

This gives

\(\frac{dy}{dx} = x^2-\frac{1}{(x+1)^3}+C.\)

Use \(\dfrac{dy}{dx}=1\) when \(x=1\):

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

So

\(C=\frac18.\)

Hence

\(\frac{dy}{dx}=x^2-\frac{1}{(x+1)^3}+\frac18.\)

(ii) Integrate again:

\(y = \int\left(x^2-\frac{1}{(x+1)^3}+\frac18\right)\,dx.\)

Therefore

\(y=\frac{x^3}{3}+\frac{1}{2(x+1)^2}+\frac{x}{8}+D.\)

Use \(y=3\) when \(x=1\):

\(3=\frac13+\frac18+\frac18+D.\)

So

\(3=\frac{7}{12}+D, \qquad D=\frac{29}{12}.\)

Thus

\(\boxed{y=\frac{x^3}{3}+\frac{1}{2(x+1)^2}+\frac{x}{8}+\frac{29}{12}}.\)

Answer: \(y=x^4+x-1\).

Integrate the derivative:

\(y=\int(4x^3+1)\,dx=x^4+x+c.\)

The curve passes through \((2,17)\), so substitute \(x=2\) and \(y=17\):

\(17=2^4+2+c=16+2+c.\)

Hence

\(c=-1.\)

Therefore the equation of the curve is

\(\boxed{y=x^4+x-1}.\)

Answer: \(y=2e^{5x}+3x+7\).

Integrate the gradient:

\(y=\int(10e^{5x}+3)\,dx.\)

This gives

\(y=2e^{5x}+3x+C.\)

The curve passes through \((0,9)\), so

\(9=2e^0+3(0)+C.\)

Thus

\(9=2+C,\)

so

\(C=7.\)

Therefore the equation of the curve is

\(\boxed{y=2e^{5x}+3x+7}.\)

Answer: \(y=x^3+\dfrac1x-x+2\).

Integrate the second derivative once:

\(\dfrac{dy}{dx}=\int\left(6x+\dfrac2{x^3}\right)\,dx.\)

This gives

\(\dfrac{dy}{dx}=3x^2-\dfrac1{x^2}+C.\)

Use \(\dfrac{dy}{dx}=1\) when \(x=1\):

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

So

\(C=-1.\)

Thus

\(\dfrac{dy}{dx}=3x^2-\dfrac1{x^2}-1.\)

Integrate again:

\(y=x^3+\dfrac1x-x+D.\)

Use \(y=3\) when \(x=1\):

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

Therefore

\(D=2.\)

Hence

\(\boxed{y=x^3+\dfrac1x-x+2}.\)

Answer: \(y=2x^{\frac32}-2x^{\frac12}-10\); tangent \(y=\dfrac{11}{2}x-20\).

The gradient of the tangent is the negative reciprocal of the gradient of the normal.

So

\(\dfrac{dy}{dx}=-\dfrac{1-3x}{\sqrt x}.\)

This simplifies to

\(\dfrac{dy}{dx}=3x^{\frac12}-x^{-\frac12}.\)

Integrate:

\(y=2x^{\frac32}-2x^{\frac12}+C.\)

The curve passes through \((1,-10)\), so

\(-10=2(1)^{\frac32}-2(1)^{\frac12}+C.\)

Thus

\(C=-10.\)

Hence the curve is

\(y=2x^{\frac32}-2x^{\frac12}-10.\)

When \(x=4\),

\(y=2(4)^{\frac32}-2(4)^{\frac12}-10=16-4-10=2.\)

The normal gradient at \(x=4\) is

\(\dfrac{\sqrt4}{1-3(4)}=\dfrac2{-11}=-\dfrac2{11}.\)

Therefore the tangent gradient is

\(\dfrac{11}{2}.\)

The tangent passes through \((4,2)\), so

\(y-2=\dfrac{11}{2}(x-4).\)

Thus

\(\boxed{y=\dfrac{11}{2}x-20}.\)