Answer: \(2(x+\frac54)^2-\frac18\), stationary point \((-\frac54,-\frac18)\), least \(p=-\frac54\).

(a) Complete the square:

\(2x^2+5x+3=2\left(x^2+\dfrac52x\right)+3.\)

Inside the bracket,

\(x^2+\dfrac52x=\left(x+\dfrac54\right)^2-\left(\dfrac54\right)^2.\)

Therefore

\(2x^2+5x+3=2\left[\left(x+\dfrac54\right)^2-\dfrac{25}{16}\right]+3.\)

Simplifying,

\(2x^2+5x+3=2\left(x+\dfrac54\right)^2-\dfrac{25}{8}+3 =2\left(x+\dfrac54\right)^2-\dfrac18.\)

Thus \(a=\dfrac54\) and \(b=-\dfrac18\).

(b) The completed square form shows that the minimum point of the parabola is

\(\left(-\dfrac54,-\dfrac18\right).\)

(c)(i) For \(\mathrm f^{-1}\) to exist, \(\mathrm f\) must be one-one on its domain. The parabola is one-one on either side of its vertex. Since the domain is \(x\geqslant p\), the least value of \(p\) that keeps only the right-hand branch is

\(p=-\dfrac54.\)

(c)(ii) With \(p=-\dfrac54\), the graph of \(y=\mathrm f(x)\) is the right-hand branch of the parabola with vertex

\(\left(-\dfrac54,-\dfrac18\right).\)

The \(y\)-intercept of \(y=\mathrm f(x)\) is found by putting \(x=0\):

\(y=3,\)

so the \(y\)-intercept is \((0,3)\).

For the \(x\)-intercepts, solve

\(2x^2+5x+3=0.\)

Factorising,

\((2x+3)(x+1)=0,\)

so \(x=-\dfrac32\) or \(x=-1\). The value \(-\dfrac32\) is not in the restricted domain \(x\geqslant-\dfrac54\), so the only \(x\)-intercept of \(\mathrm f\) is

\((-1,0).\)

The graph of \(y=\mathrm f^{-1}(x)\) is the reflection of \(y=\mathrm f(x)\) in the line \(y=x\). Therefore the intercept \((-1,0)\) reflects to \((0,-1)\), and the intercept \((0,3)\) reflects to \((3,0)\). The vertex reflects to

\(\left(-\dfrac18,-\dfrac54\right).\)

Answer: range \(\mathrm f(x)\gt2\); \(\mathrm f^{-1}(x)=-\frac13\ln(x-2)\); asymptotes \(y=2\) and \(x=2\); \(\mathrm{gf}(x)=12\) gives \(x=-\frac13\ln2\).

We start with the main method. Use the measurements and relationships shown in the diagram, then translate them into algebraic or trigonometric equations. Use the fact that the graph of an inverse function is the reflection in the line \(y=x\).

(a) Since \(e^{-3x}\gt0\) for all real \(x\),

\(\mathrm f(x)=2+e^{-3x}\gt2.\)

So the range is

\(\mathrm f(x)\gt2.\)

(b) Let

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

Then

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

Taking natural logarithms gives

\(\ln(y-2)=-3x.\)

So

\(x=-\frac13\ln(y-2).\)

Therefore

\(\mathrm f^{-1}(x)=-\frac13\ln(x-2),\quad x\gt2.\)

(c) The graph of \(y=\mathrm f(x)\) has \(y\)-intercept

\(2+e^0=3,\)

so it passes through \((0,3)\). Its horizontal asymptote is

\(y=2.\)

The graph of \(y=\mathrm f^{-1}(x)\) is the reflection of \(y=\mathrm f(x)\) in the line \(y=x\). It has \(x\)-intercept \((3,0)\) and vertical asymptote

\(x=2.\)

(d) The equation \(\mathrm{gf}(x)=12\) means

\((\mathrm f(x))^{3/2}+4=12.\)

So

\((2+e^{-3x})^{3/2}=8.\)

Raise both sides to the power \(\frac23\):

\(2+e^{-3x}=4.\)

Thus

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

Taking natural logarithms,

\(-3x=\ln2,\)

so

\(x=-\frac13\ln2.\)

This completes the solution and gives the required result.

Answer: (a)(i) domain \(0\leq x\leq\frac94\), range \(0\leq f^{-1}(x)\leq3\); (a)(ii) \(x=0\) or \(x=\frac85\); (b)(i) \(g^{-1}(x)=\sqrt[3]{\frac{x^3-3}{8}}\); (b)(ii) \(k=0\); (b)(iii) \(gh(x)=\sqrt[3]{8e^{12x}+3}\).

For an inverse function, interchange \(x\) and \(y\), then rearrange to make \(y\) the subject. The domain of the original function becomes the range of the inverse.

(a)(i) The domain of \(f^{-1}\) is the range of \(f\). Since

\(f(0)=0\)

and

\(f(3)=\frac{9}{\sqrt{16}}=\frac94,\)

the domain of \(f^{-1}\) is

\(0\leq x\leq\frac94.\)

The range of \(f^{-1}\) is the domain of \(f\), so

\(0\leq f^{-1}(x)\leq3.\)

(a)(ii) Solve

\(\frac{3x}{\sqrt{5x+1}}=x.\)

One solution is \(x=0\). If \(x\ne0\), divide by \(x\):

\(\frac{3}{\sqrt{5x+1}}=1.\)

Hence

\(\sqrt{5x+1}=3.\)

Squaring gives

\(5x+1=9,\)

so

\(x=\frac85.\)

Thus

\(x=0\quad\text{or}\quad x=\frac85.\)

(a)(iii) The graph of \(y=f^{-1}(x)\) is the reflection of the graph of \(y=f(x)\) in the line \(y=x\). It passes through \((0,0)\) and \(\left(\frac94,3\right)\), and it intersects the graph of \(y=f(x)\) at \((0,0)\) and \(\left(\frac85,\frac85\right)\).

(b)(i) Let

\(y=\sqrt[3]{8x^3+3}.\)

Then

\(y^3=8x^3+3.\)

So

\(x^3=\frac{y^3-3}{8}.\)

Therefore

\(x=\sqrt[3]{\frac{y^3-3}{8}}.\)

Replacing \(y\) by \(x\),

\(g^{-1}(x)=\sqrt[3]{\frac{x^3-3}{8}}.\)

(b)(ii) For \(gh(x)\) to be formed, the output of \(h\) must be in the domain of \(g\). Since the domain of \(g\) is \(x\geq1\), we need

\(e^{4x}\geq1.\)

This is true when \(x\geq0\), so the least possible value is

\(k=0.\)

(b)(iii) Now

\(gh(x)=g(h(x))=g(e^{4x}).\)

Therefore

\(gh(x)=\sqrt[3]{8(e^{4x})^3+3} =\sqrt[3]{8e^{12x}+3}.\)

Answer: (a)(i) \(f(x)\leq -1\); (a)(ii) \(x=-2\); (a)(iii) \(\frac{\pi}{2}\lt x\lt \frac{3\pi}{2}\); (a)(iv) \(x=\frac{5\pi}{6},\frac{7\pi}{6}\).

For an inverse function, interchange \(x\) and \(y\), then rearrange to make \(y\) the subject. The domain of the original function becomes the range of the inverse.

(a)(i) For

\(\frac{\pi}{2}\lt x\lt \frac{3\pi}{2},\)

we have \(\cos x\lt 0\). Therefore

\(\operatorname{sec} x=\frac{1}{\cos x}\leq -1.\)

So the range of \(f\) is

\(f(x)\leq -1.\)

(a)(ii) If

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

then

\(x=f\left(\frac{2\pi}{3}\right).\)

Hence

\(x=\operatorname{sec}\frac{2\pi}{3}.\)

Since

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

we get

\(x=-2.\)

(a)(iii) Since \(g\) is defined for all real inputs, the domain of \(gf\) is just the domain of \(f\):

\(\frac{\pi}{2}\lt x\lt \frac{3\pi}{2}.\)

(a)(iv) Now

\(gf(x)=g(f(x))=3\bigl((\operatorname{sec} x)^2-1\bigr).\)

Using \(\operatorname{sec}^2x-1=\tan^2x\),

\(gf(x)=3\tan^2x.\)

So \(gf(x)=1\) gives

\(3\tan^2x=1.\)

Therefore

\(\tan^2x=\frac13.\)

In the interval \(\frac{\pi}{2}\lt x\lt \frac{3\pi}{2}\), the solutions are

\(x=\frac{5\pi}{6} \quad\text{and}\quad x=\frac{7\pi}{6}.\)

(b) For

\(h(x)=\ln(4-x),\)

the vertical asymptote is

\(x=4.\)

The intercepts of \(y=h(x)\) are:

\(y\text{-intercept: }(0,\ln4),\)

and

\(x\text{-intercept: }(3,0).\)

To find the inverse, let

\(y=\ln(4-x).\)

Then

\(e^y=4-x,\)

so

\(x=4-e^y.\)

Thus

\(h^{-1}(x)=4-e^x.\)

The graph of \(y=h^{-1}(x)\) has horizontal asymptote

\(y=4.\)

Its intercepts are

\((0,3) \quad\text{and}\quad (\ln4,0).\)

The two graphs are reflections of each other in the line \(y=x\).

Answer: (a)(i) \(a=-\frac13\), (ii) \(f\geq-4\); (b) \(x=\sqrt{\frac{e^2-5}{2}}\).

For an inverse function, interchange \(x\) and \(y\), then rearrange to make \(y\) the subject. The domain of the original function becomes the range of the inverse.

(a)(i) The function is

\(f(x)=(3x+1)^2-4\).

This is a quadratic with vertex at

\(3x+1=0\),

so

\(x=-\frac13\).

For \(f^{-1}\) to exist on the domain \(x\geq a\), the function must be one-to-one on that domain. The least possible value is therefore the \(x\)-coordinate of the vertex:

\(a=-\frac13\).

(a)(ii) At the vertex,

\(f\left(-\frac13\right)=-4\).

Since the parabola opens upwards and the domain starts at the vertex, the range is

\(f(x)\geq-4\).

(a)(iii) The graph of \(y=f(x)\) is the right-hand half of the parabola with vertex \(\left(-\frac13,-4\right)\).

It passes through the \(y\)-axis at

\(f(0)=1-4=-3\),

so it passes through \((0,-3)\).

It meets the \(x\)-axis when

\((3x+1)^2-4=0\).

So \((3x+1)^2=4\), giving \(3x+1=2\) on the chosen domain, and hence \(x=\frac13\).

Thus \(y=f(x)\) passes through \(\left(\frac13,0\right)\).

The graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\).

So it passes through \((-3,0)\) and \(\left(0,\frac13\right)\), and its vertex is \(\left(-4,-\frac13\right)\).

(b) Since \(h(x)=3x-2\),

\(hg(x)=h(g(x))=3\ln(2x^2+5)-2\).

Set this equal to 4:

\(3\ln(2x^2+5)-2=4\).

So

\(3\ln(2x^2+5)=6\),

and hence

\(\ln(2x^2+5)=2\).

Therefore

\(2x^2+5=e^2\).

So

\(x^2=\frac{e^2-5}{2}\).

Since \(x\geq0\),

\(x=\sqrt{\frac{e^2-5}{2}}\).

Answer: (a)(i) \(a=-\frac25\). (ii) Range: \(\mathbb{R}\). (iii) \(\displaystyle f^{-1}(x)=\frac{e^x-2}{5}\), domain \(\mathbb{R}\). (b) \(x=64\).

Answer: (a)(i) \(a=-\frac25\). (ii) Range: \(\mathbb{R}\). (iii) \(\displaystyle f^{-1}(x)=\frac{e^x-2}{5}\), domain \(\mathbb{R}\). (b) \(x=64\).

(a)(i) For \(f(x)=\ln(5x+2)\), the argument must be positive:

\(5x+2\gt 0\).

Thus \(x\gt -\frac25\), so the smallest possible value of \(a\) is

\(a=-\frac25\).

(a)(ii) The logarithm function can take all real values, so the range of \(f\) is \(\mathbb{R}\).

(a)(iii) Let \(y=\ln(5x+2)\). Then

\(e^y=5x+2\),

so

\(\displaystyle x=\frac{e^y-2}{5}\).

Therefore

\(\displaystyle f^{-1}(x)=\frac{e^x-2}{5}\).

The domain of \(f^{-1}\) is the range of \(f\), so the domain is \(\mathbb{R}\).

(a)(iv) The graph of \(y=f(x)\) has vertical asymptote \(x=-\frac25\), \(x\)-intercept \(\left(-\frac15,0\right)\), and \(y\)-intercept \((0,\ln2)\).

The graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\). Its intercepts are \((\ln2,0)\) and \(\left(0,-\frac15\right)\).

(b) Since \(g(x)=\sqrt{x}-4\),

\(g^2(x)=g(g(x))=\sqrt{\sqrt{x}-4}-4\).

Set this equal to \(-2\):

\(\sqrt{\sqrt{x}-4}-4=-2\).

Then

\(\sqrt{\sqrt{x}-4}=2\).

Squaring gives

\(\sqrt{x}-4=4\),

so \(\sqrt{x}=8\), and therefore

\(x=64\).

Answer: (a)(i) \(3\leq x\lt 5\); (a)(ii) \(x=4\); (b) \(\displaystyle f^{-1}g(x)=\ln\left(\frac{2-2x}{2-5x}\right)\).

Answer: (a)(i) \(3\leq x\lt 5\); (a)(ii) \(x=4\); (b) \(\displaystyle f^{-1}g(x)=\ln\left(\frac{2-2x}{2-5x}\right)\).

(a)(i) Since \(x\geq0\), we have \(0\lt \mathrm e^{-x}\leq1\).

Therefore

\(3\leq 5-2\mathrm e^{-x}\lt 5.\)

The range of \(f\) is \(3\leq y\lt 5\), so the domain of \(f^{-1}\) is

\(3\leq x\lt 5.\)

(a)(ii) If \(f^{-1}(x)=\sqrt{5x-4}\), apply \(f\) to both sides:

\(x=f(\sqrt{5x-4}).\)

It is quicker to note that \(f^{-1}(x)=u\) means \(x=f(u)\). The equation given also says \(u=\sqrt{5x-4}\). Thus the mark scheme equation reduces to

\(x=\sqrt{5x-4}.\)

Squaring gives

\(x^2=5x-4,\)

so

\(x^2-5x+4=0.\)

Factorising,

\((x-1)(x-4)=0.\)

The domain of \(f^{-1}\) is \(3\leq x\lt 5\), so reject \(x=1\). Hence

\(x=4.\)

(a)(iii) The graph of \(y=f(x)\) starts at \((0,3)\), increases, and has horizontal asymptote \(y=5\).

The graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\). It starts at \((3,0)\), increases, and has vertical asymptote \(x=5\).

(b) To find \(f^{-1}\), let

\(y=5-2\mathrm e^{-x}.\)

Then

\(2\mathrm e^{-x}=5-y,\)

so

\(\mathrm e^{-x}=\frac{5-y}{2}.\)

Taking logarithms,

\(-x=\ln\left(\frac{5-y}{2}\right),\)

and hence

\(f^{-1}(y)=\ln\left(\frac{2}{5-y}\right).\)

Now substitute \(g(x)=\frac{3}{1-x}\):

\(f^{-1}g(x)=\ln\left(\frac{2}{5-\frac{3}{1-x}}\right).\)

Simplify the denominator:

Therefore

\(f^{-1}g(x)=\ln\left(\frac{2}{(2-5x)/(1-x)}\right) =\ln\left(\frac{2-2x}{2-5x}\right).\)

Answer: \(g^{-1}(x)=4-\sqrt{x-1}\), domain \(x\geq1\), range \(g^{-1}(x)\leq4\); \(fg(x)=\ln(2x^2-16x+35)\).

(a) The graph of

\(f(x)=\ln(2x+1),\qquad x\geq0,\)

starts at

\(f(0)=\ln1=0,\)

and then increases, with gradually decreasing gradient.

The graph of \(y=f^{-1}(x)\) is the reflection of the graph of \(y=f(x)\) in the line \(y=x\). It also passes through \((0,0)\), but has the corresponding inverse shape.

(b)(i) Let

\(y=(x-4)^2+1.\)

Then

\(y-1=(x-4)^2.\)

Since \(x\leq4\), we must take the negative square root:

\(x-4=-\sqrt{y-1}.\)

So

\(x=4-\sqrt{y-1}.\)

Therefore

\(g^{-1}(x)=4-\sqrt{x-1}.\)

The range of \(g\) is \(g(x)\geq1\), so the domain of \(g^{-1}\) is

\(x\geq1.\)

The domain of \(g\) is \(x\leq4\), so the range of \(g^{-1}\) is

\(g^{-1}(x)\leq4.\)

(b)(ii) We have

\(fg(x)=f(g(x)).\)

Since \(f(x)=\ln(2x+1)\),

\(f(g(x))=\ln(2g(x)+1).\)

Substitute \(g(x)=(x-4)^2+1\):

\(fg(x)=\ln\left(2\left[(x-4)^2+1\right]+1\right).\)

Simplify the expression inside the logarithm:

\(2\left[(x-4)^2+1\right]+1 =2(x^2-8x+16)+3 =2x^2-16x+35.\)

Therefore

\(fg(x)=\ln(2x^2-16x+35).\)

(b)(iii) The function \(gf\) would mean \(g(f(x))\). For this to exist for the whole domain of \(f\), every output of \(f\) must lie in the domain of \(g\).

But \(f(x)=\ln(2x+1)\) is unbounded above, while the domain of \(g\) is \(x\leq4\). Some values in the range of \(f\) are greater than \(4\), so they are outside the domain of \(g\).

Therefore \(gf\) does not exist.

Answer: \(\mathrm{f}^{-1}\) exists because \(\mathrm{f}\) is one-one for \(x\geq-1\). The vertices are \((-1,-4)\) and \((-4,-1)\).

(a) Complete the square:

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

The vertex of \(y=\mathrm{f}(x)\) is \((-1,-4)\). Since the domain is restricted to \(x\geq-1\), only the right-hand branch of the parabola is used. On this interval the function is one-one, so \(\mathrm{f}^{-1}\) exists.

(b) The graph of \(y=\mathrm{f}(x)\) has vertex

\((-1,-4).\)

Its \(y\)-intercept is

\(\mathrm{f}(0)=-3,\)

and its \(x\)-intercept in the domain is

\(x^2+2x-3=0 \quad\Rightarrow\quad (x+3)(x-1)=0,\)

so \(x=1\).

The graph of \(y=\mathrm{f}^{-1}(x)\) is the reflection of \(y=\mathrm{f}(x)\) in the line \(y=x\). Therefore its vertex is

\((-4,-1).\)

The intercepts are also swapped: it has \(x\)-intercept \(-3\) and \(y\)-intercept \(1\).

Answer: Range of \(f\): \(y\gt 1\); range of \(g\): \(\mathbb{R}\); \(fg^2(0)=3e^4+1\approx165\). Intercepts: \((0,4)\) and \((4,0)\).

Since \(e^{2x}\gt 0\) for all real \(x\),

\(3e^{2x}+1\gt 1.\)

So the range of \(f\) is

\(y\gt 1.\)

The function \(g(x)=x+1\) is linear with domain \(\mathbb{R}\), so its range is

\(\mathbb{R}.\)

Next, \(g^2(0)\) means \(g(g(0))\). Since

\(g(0)=1,\qquad g(1)=2,\)

we have

\(fg^2(0)=f(2)=3e^4+1\approx165.\)

For the sketch, \(y=f(x)\) has horizontal asymptote \(y=1\) and meets the y-axis at

\(f(0)=4,\)

so the point is \((0,4)\).

The graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\). Therefore it meets the x-axis at the reflected point

\((4,0).\)

Answer: (i) \(f(x)=2\left(x-\dfrac32\right)^2+\dfrac12\). (ii) The graphs are reflections in the line \(y=x\). (iii) \(f^{-1}(x)=\dfrac32-\sqrt{\dfrac{x-\frac12}{2}}\), with domain \(x\geqslant \dfrac12\).

(i) Complete the square:

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

Now

\(x^2-3x=\left(x-\frac32\right)^2-\frac94,\)

so

\(f(x)=2\left[\left(x-\frac32\right)^2-\frac94\right]+5 =2\left(x-\frac32\right)^2-\frac92+5 =2\left(x-\frac32\right)^2+\frac12.\)

(ii) The graph of \(y=f(x)\) is the left-hand branch of the parabola with vertex \(\left(\frac32,\frac12\right)\), because the domain is \(x\leqslant\frac32\).

The graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\). In particular, the vertex \(\left(\frac32,\frac12\right)\) reflects to \(\left(\frac12,\frac32\right)\).

(iii) Start from

\(y=2\left(x-\frac32\right)^2+\frac12.\)

Swap \(x\) and \(y\):

\(x=2\left(y-\frac32\right)^2+\frac12.\)

Then

\(\frac{x-\frac12}{2}=\left(y-\frac32\right)^2.\)

Taking square roots gives

\(y-\frac32=\pm\sqrt{\frac{x-\frac12}{2}}.\)

Because the original domain is \(x\leqslant\frac32\), the inverse must use the branch with \(y\leqslant\frac32\). So we take the negative root:

\(f^{-1}(x)=\frac32-\sqrt{\frac{x-\frac12}{2}}.\)

The domain of \(f^{-1}\) is the range of \(f\), which is

\(x\geqslant\frac12.\)

Answer: (i) \(B\) and \(C\). (ii) \(B\) only.

(i) A graph represents a function if every vertical line meets it at most once.

Graph \(A\) is a circle, so many vertical lines cut it twice. It is not a function.

Graph \(B\) passes the vertical line test, so it is a function.

Graph \(C\) also passes the vertical line test, so it is a function.

Graph \(D\) has an upper branch and a lower branch, so some vertical lines meet it twice. It is not a function.

Therefore the graphs that represent functions are

\(\boxed{B\text{ and }C}.\)

(ii) A function has an inverse only if it is one-to-one, which means it also passes the horizontal line test.

Graph \(B\) is one-to-one, so it has an inverse.

Graph \(C\) is not one-to-one because many horizontal lines meet it twice.

Hence the only graph that represents a function with an inverse is

\(\boxed{B}.\)

Answer: (i) \(f^{-1}(x)=\left(\dfrac{x}{9}\right)^2+1\), domain \(x\gt 0\). (ii) \(45\sqrt2\). (iii) \(x=\dfrac85\).

(i) Let

\(y=9\sqrt{x-1}.\)

Then

\(\frac{y}{9}=\sqrt{x-1}.\)

Square both sides:

\(\left(\frac{y}{9}\right)^2=x-1.\)

So

\(x=\left(\frac{y}{9}\right)^2+1.\)

Swap \(x\) and \(y\):

\(f^{-1}(x)=\left(\frac{x}{9}\right)^2+1.\)

Since \(x\gt 1\) for \(f\), we have \(\sqrt{x-1}\gt 0\), so the range of \(f\) is \(y\gt 0\). Therefore the domain of \(f^{-1}\) is

\(x\gt 0.\)

(ii) First find \(g(7)\):

\(g(7)=7^2+2=51.\)

Therefore

\(fg(7)=f(51)=9\sqrt{51-1}=9\sqrt{50}=45\sqrt2.\)

(iii) We have

\(gf(x)=g(f(x))=(9\sqrt{x-1})^2+2=81(x-1)+2=81x-79.\)

So the equation becomes

\(81x-79=5x^2+83x-95.\)

Rearranging,

\(5x^2+2x-16=0.\)

Factorise:

\((5x-8)(x+2)=0.\)

Hence

\(x=\frac85 \quad\text{or}\quad x=-2.\)

But the domain is \(x\gt 1\), so only

\(\boxed{x=\frac85}\)

is valid.