Past Exam Questions

Answer: \(\mathrm f^{-1}(x)=-\sqrt{\ln x-3}\), domain \(x\gt \mathrm e^3\), range \(\mathrm f^{-1}(x)\lt 0\), and \(\mathrm g(x)=-\sqrt{2x-3}\).

We start with the main method. Check whether the function is one-to-one on the stated domain before finding an inverse.

The function \(\mathrm f\) is defined by

\(\mathrm f(x)=\mathrm e^{x^2+3}\quad\text{for }x\lt 0.\)

On \(x\lt 0\), as \(x\) increases, \(x^2\) decreases. Hence \(\mathrm f\) is one-one, so \(\mathrm f^{-1}\) exists.

To find the inverse, let

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

Then

\(\ln y=x^2+3.\)

So

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

Since the original domain is \(x\lt 0\), take the negative square root:

\(x=-\sqrt{\ln y-3}.\)

Therefore

\(\mathrm f^{-1}(x)=-\sqrt{\ln x-3}.\)

The domain of \(\mathrm f^{-1}\) is the range of \(\mathrm f\). Since \(x\lt 0\), \(x^2\gt 0\), so \(x^2+3\gt 3\), and therefore

\(x\gt \mathrm e^3.\)

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

\(\mathrm f^{-1}(x)\lt 0.\)

Now use \(\mathrm{fg}(x)=\mathrm e^{2x}\). This means

\(\mathrm f(\mathrm g(x))=\mathrm e^{2x}.\)

Apply \(\mathrm f^{-1}\) to both sides:

\(\mathrm g(x)=\mathrm f^{-1}(\mathrm e^{2x}).\)

Hence

\(\mathrm g(x)=-\sqrt{\ln(\mathrm e^{2x})-3}=-\sqrt{2x-3}.\)

This completes the solution and gives the required result.

Answer: (a) \(x=\ln\left(\frac3{\sqrt2}-1\right)\); (b) \(\mathrm f^{-1}\) does not exist; \(\mathrm g^{-1}(x)=\ln(x-1)\), domain \(x\gt1\).

We start with the main method. Check whether the function is one-to-one on the stated domain before finding an inverse.

(a) Since \(\mathrm{fg}(x)=\mathrm f(\mathrm g(x))\),

\(\mathrm{fg}(x)=2(\mathrm e^x+1)^2-1.\)

Set this equal to \(8\):

\(2(\mathrm e^x+1)^2-1=8.\)

Then

\(2(\mathrm e^x+1)^2=9,\)

so

\((\mathrm e^x+1)^2=\frac92.\)

Because \(\mathrm e^x+1\gt0\),

\(\mathrm e^x+1=\frac3{\sqrt2}.\)

Therefore

\(\mathrm e^x=\frac3{\sqrt2}-1.\)

So

\(x=\ln\left(\frac3{\sqrt2}-1\right).\)

(b) The function \(\mathrm f(x)=2x^2-1\) is not one-to-one for all real values of \(x\). For example, \(\mathrm f(1)=\mathrm f(-1)\).

Therefore \(\mathrm f^{-1}\) does not exist on the stated domain.

For \(\mathrm g\), let

\(y=\mathrm e^x+1.\)

Then

\(y-1=\mathrm e^x.\)

Taking natural logarithms gives

\(x=\ln(y-1).\)

So

\(\mathrm g^{-1}(x)=\ln(x-1).\)

Since the input to the logarithm must be positive, the domain is

\(x\gt1.\)

This completes the solution and gives the required result.

Answer: domain of \(f^{-1}\): \(0\leqslant x\leqslant\dfrac{2\sqrt2}{3}\); range of \(f^{-1}\): \(\dfrac12\leqslant f^{-1}(x)\leqslant\dfrac32\); \(f^{-1}(x)=\sqrt{\dfrac1{4-4x^2}}\); \(gf(x)=e^{1-\frac1{4x^2}}\).

(a)(i) The domain of \(f\) is

\(\frac12\leqslant x\leqslant\frac32.\)

From the graph and the formula, \(f\) is increasing on this interval.

At \(x=\frac12\),

\(f\left(\frac12\right)=0.\)

At \(x=\frac32\),

So the range of \(f\), and hence the domain of \(f^{-1}\), is

\(0\leqslant x\leqslant\frac{2\sqrt2}{3}.\)

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

\(\frac12\leqslant f^{-1}(x)\leqslant\frac32.\)

(a)(ii) Let

\(y=\frac{\sqrt{4x^2-1}}{2x}.\)

Then

\(2xy=\sqrt{4x^2-1}.\)

Squaring both sides,

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

Rearrange:

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

Hence

\(x^2=\frac1{4(1-y^2)}=\frac1{4-4y^2}.\)

Since \(x\gt 0\),

\(x=\sqrt{\frac1{4-4y^2}}.\)

Interchanging \(x\) and \(y\),

\(f^{-1}(x)=\sqrt{\frac1{4-4x^2}}.\)

(b) Since \(g(x)=e^{x^2}\),

\(gf(x)=g(f(x))=e^{(f(x))^2}.\)

Now

So

\((f(x))^2=\frac{4x^2-1}{4x^2} =1-\frac1{4x^2}.\)

Therefore

\(gf(x)=e^{1-\frac1{4x^2}},\)

which is of the required form with \(a=1\) and \(b=4\).

Answer: \(\mathrm{fg}(x)=\frac{2}{3x}-\frac{x}{3}\); range of \(\mathrm f^{-1}\) is \(x\gt 0\); \(\mathrm f^{-1}(x)=\frac{3x+\sqrt{9x^2+8}}{4}\).

(a) Since \(\mathrm g(x)=\frac1x\),

Substitute \(\frac1x\) into \(\mathrm f\):

So

Therefore

\(\mathrm{fg}(x)=\frac{2}{3x}-\frac{x}{3}.\)

(b)(i) The domain of \(\mathrm f\) is \(x\gt 0\). Therefore the range of \(\mathrm f^{-1}\) is

\(x\gt 0.\)

(b)(ii) Let

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

Then

\(3xy=2x^2-1.\)

Rearrange as a quadratic in \(x\):

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

Using the quadratic formula,

\(x=\frac{3y\pm\sqrt{9y^2+8}}{4}.\)

Since \(x\gt 0\), take the positive square-root branch:

\(x=\frac{3y+\sqrt{9y^2+8}}{4}.\)

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

\(\mathrm f^{-1}(x)=\frac{3x+\sqrt{9x^2+8}}{4}.\)

Answer: \(\mathrm f(x)=15-2(x+1)^2\), range \(\mathrm f\leq15\); domain of \(\mathrm g^{-1}\): \(x\geq\sqrt2\), range: \(x\geq1\); \(\mathrm g^{-1}(x)=-1+\sqrt{x^2+2}\).

(a)(i) Complete the square:

\(\mathrm f(x)=13-4x-2x^2.\)

Factor out \(-2\) from the quadratic terms:

\(\mathrm f(x)=-2(x^2+2x)+13.\)

Since

\(x^2+2x=(x+1)^2-1,\)

we get

\(\mathrm f(x)=-2\left((x+1)^2-1\right)+13.\)

So

\(\mathrm f(x)=15-2(x+1)^2.\)

(a)(ii) Since \((x+1)^2\geq0\),

\(-2(x+1)^2\leq0.\)

Therefore

\(\mathrm f(x)\leq15.\)

(b)(i) The domain of \(\mathrm g\) is \(x\geq1\). Hence the range of \(\mathrm g^{-1}\) is

\(x\geq1.\)

Also

\(\mathrm g(1)=\sqrt{1+2-1}=\sqrt2,\)

and \(\mathrm g\) is increasing for \(x\geq1\). So the range of \(\mathrm g\), and hence the domain of \(\mathrm g^{-1}\), is

\(x\geq\sqrt2.\)

(b)(ii) Let

\(y=\sqrt{x^2+2x-1}.\)

Then

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

Rearrange as a quadratic in \(x\):

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

Using the quadratic formula,

\(x=\frac{-2\pm\sqrt{4+4(1+y^2)}}{2}.\)

So

\(x=-1\pm\sqrt{y^2+2}.\)

Since the range of \(\mathrm g^{-1}\) is \(x\geq1\), choose the positive branch:

\(x=-1+\sqrt{y^2+2}.\)

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

\(\mathrm g^{-1}(x)=-1+\sqrt{x^2+2}.\)

Answer: \(g^{-1}(x)=\frac{1}{x-3}\), domain \(3\lt x\leq4\), range \(x\geq1\).

(a) Let

\(y=3+\frac1x.\)

Then

\(y-3=\frac1x,\)

so

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

Therefore

\(\boxed{g^{-1}(x)=\frac{1}{x-3}}.\)

For \(x\geq1\), the value of \(\frac1x\) satisfies \(0\lt \frac1x\leq1\). Hence

\(3\lt g(x)\leq4.\)

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

\(\boxed{3\lt x\leq4},\)

and the range of \(g^{-1}\) is

\(\boxed{x\geq1}.\)

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

The graph of \(y=h(x)\) begins to the right of \(x=\frac23\), is increasing, and is concave down. Since \(y=h(x)\) intersects \(y=x\) at two distinct points, the graph of \(y=h^{-1}(x)\) passes through those same two points after reflection.

Answer: range \(f\gt9\); \(f^{-1}(x)=\dfrac{\sqrt x-3}{2}\), domain \(x\gt9\); \(x=e^2-4\).

(a) Since \(x\gt0\),

\(2x+3\gt3.\)

Therefore

\((2x+3)^2\gt9.\)

So the range is

\(f\gt9.\)

(b) On the restricted domain \(x\gt0\), the function \(f(x)=(2x+3)^2\) is one-to-one. Therefore \(f\) has an inverse.

(c) Let

\(y=(2x+3)^2.\)

Since \(x\gt0\), \(2x+3\gt0\), so

\(\sqrt y=2x+3.\)

Hence

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

Therefore

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

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

\(x\gt9.\)

(e) We need

\(f(g(x))=49.\)

Since \(g(x)=\ln(x+4)\),

\(\left(2\ln(x+4)+3\right)^2=49.\)

Because \(x\gt0\), we have \(\ln(x+4)\gt0\), so \(2\ln(x+4)+3\) is positive. Therefore

\(2\ln(x+4)+3=7.\)

Thus

\(\ln(x+4)=2.\)

So

\(x+4=e^2,\)

and hence

\(x=e^2-4.\)

Answer: \(g^2(x)=6(6x^4+5)^4+5\), domain \(\mathbb R\); \(k=0\); \(h^{-1}(x)=-\sqrt[4]{\dfrac{x-5}{6}}\); range of \(p\) is \(p(x)\gt 2\).

The function \(g\) is defined for every real input, and each input has exactly one output. However, it is not one-to-one because

\(g(x)=g(-x).\)

For example, \(g(1)=g(-1)\). Therefore \(g\) does not have an inverse on its full domain.

Now

\(g^2(x)=g(g(x)).\)

Since \(g(x)=6x^4+5\),

\(g^2(x)=6(6x^4+5)^4+5.\)

The domain is all real \(x\).

For \(h(x)=6x^4+5\) to have an inverse on \(x\leq k\), the largest possible interval of this form on which the function is one-to-one is \(x\leq0\). Therefore

\(k=0.\)

To find \(h^{-1}\), write

\(y=6x^4+5,\qquad x\leq0.\)

Then

\(x^4=\frac{y-5}{6}.\)

Since \(x\leq0\), take the negative fourth root:

\(x=-\sqrt[4]{\frac{y-5}{6}}.\)

Hence

\(h^{-1}(x)=-\sqrt[4]{\frac{x-5}{6}}.\)

For \(p(x)=3e^x+2\), since \(e^x\gt 0\),

\(p(x)\gt 2.\)

So the range of \(p\) is \(p(x)\gt 2\).

The graph of \(y=p(x)\) is an increasing exponential curve with horizontal asymptote \(y=2\). It crosses the \(y\)-axis at

\(p(0)=5,\)

so its \(y\)-intercept is \((0,5)\), and it has no \(x\)-intercept.

The graph of \(y=p^{-1}(x)\) is the reflection of \(y=p(x)\) in the line \(y=x\). It has vertical asymptote \(x=2\), crosses the \(x\)-axis at \((5,0)\), and has no \(y\)-intercept.

The two graphs are reflections in \(y=x\), but they do not intersect. Therefore the equation

\(p(x)=p^{-1}(x)\)

has no solutions.

Answer: Range of \(g\): \(y\geq1\). Also \(f^{-1}g(\sqrt{62})=\ln5\), and \(x=\ln2\).

Since \(g(x)=2x^2+1\) and \(x\geq0\), the smallest value occurs at \(x=0\). Therefore

\(g(x)\geq1,\)

so the range is

\(\boxed{y\geq1}.\)

For \(f(x)=e^{3x}\), the inverse is found by writing

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

Then \(\ln y=3x\), so

\(f^{-1}(y)=\frac13\ln y.\)

Now

\(g(\sqrt{62})=2(\sqrt{62})^2+1=125.\)

Hence

\(f^{-1}g(\sqrt{62})=\frac13\ln125.\)

Since \(125=5^3\),

\(\frac13\ln125=\frac13\ln(5^3)=\boxed{\ln5}.\)

Next,

\(f'(x)=3e^{3x}.\)

Also,

\(g'(x)=4x,\qquad g''(x)=4.\)

The equation \(f'(x)=6g''(x)\) gives

\(3e^{3x}=6(4)=24.\)

Thus

\(e^{3x}=8,\)

so

\(3x=\ln8=\ln(2^3),\)

and therefore

\(\boxed{x=\ln2}.\)

For the sketch, \(y=g(x)=2x^2+1\) is the right-hand half of an upward parabola, starting at \((0,1)\). Its inverse \(y=g^{-1}(x)\) is the reflection in the line \(y=x\), starting at \((1,0)\).

Answer: range \(g\gt -9\); domain \(x\gt 1\); \(x=\dfrac{4+3\sqrt6}{10}\); reflection in \(y=x\); \(h^{-1}(x)=-\sqrt{x^2+1}\).

For \(g(x)=4x^2-9\) with \(x\gt 0\), the values of \(4x^2\) are positive and can get arbitrarily close to 0. Therefore

\(\boxed{\text{range of }g\text{ is }y\gt -9}.\)

For \(gf\), we first apply \(f\), then \(g\). Since \(f\) is defined for \(x\gt 1\), and

\(f(x)=5x-2\gt 3\)

when \(x\gt 1\), the output of \(f\) is always in the domain of \(g\). Hence the domain of \(gf\) is

\(\boxed{x\gt 1}.\)

Now

\(gf(x)=g(f(x))=4(5x-2)^2-9.\)

Set this equal to 45:

\(4(5x-2)^2-9=45.\)

Then

\(4(5x-2)^2=54,\)

so

\((5x-2)^2=\frac{27}{2}.\)

Thus

\(5x-2=\pm\frac{3\sqrt6}{2}.\)

So

\(x=\frac{4\pm3\sqrt6}{10}.\)

Only the positive-sign value satisfies \(x\gt 1\), so

\(\boxed{x=\frac{4+3\sqrt6}{10}}.\)

The graphs of \(y=h(x)\) and \(y=h^{-1}(x)\) are reflections of each other in the line

\(\boxed{y=x}.\)

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

\(y=\sqrt{x^2-1},\qquad x\leq-1.\)

Then

\(y^2=x^2-1,\)

so

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

Since the original \(x\leq-1\), choose the negative root:

\(x=-\sqrt{y^2+1}.\)

Therefore

\(\boxed{h^{-1}(x)=-\sqrt{x^2+1}}.\)

Answer: \(f\) is not one-to-one; \(f^2(x)=\sqrt{x^2+2}\); one possible value is \(k=0\); \(g^{-1}(x)=\sqrt{x^2-1}\). Intercepts: \((0,6)\) for \(h\), and \((6,0)\) for \(h^{-1}\).

The graph of \(f(x)=\sqrt{1+x^2}\) is symmetric about the \(y\)-axis. A horizontal line above \(y=1\) cuts the graph twice.

Therefore \(f\) is not one-to-one, so it does not have an inverse on the full real domain.

Now

\(f^2(x)=f(f(x)).\)

Since \(f(x)=\sqrt{1+x^2}\),

\(f^2(x)=\sqrt{1+\left(\sqrt{1+x^2}\right)^2}=\sqrt{x^2+2}.\)

For \(g\) to have an inverse, restrict the domain to one side of the turning point. One possible value is

\(k=0.\)

Then \(g\) is defined for \(x\gt 0\), where it is increasing.

To find the inverse, write

\(y=\sqrt{1+x^2}.\)

Square both sides:

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

Since \(x\gt 0\),

\(x=\sqrt{y^2-1}.\)

So

\(g^{-1}(x)=\boxed{\sqrt{x^2-1}}.\)

For \(h(x)=4e^x+2\), the graph is an increasing exponential curve with horizontal asymptote \(y=2\). It crosses the \(y\)-axis at

\(h(0)=4e^0+2=6,\)

so the intercept is \((0,6)\).

The inverse graph is the reflection of \(y=h(x)\) in the line \(y=x\). Therefore it crosses the \(x\)-axis at \((6,0)\) and has vertical asymptote \(x=2\).

Answer: (i) \(\mathrm f^{-1}(x)=\dfrac{5x+1}{2x}\). (ii) \(x\gt0\). (iii) \(\mathrm f^2(x)=\dfrac{2x-5}{-10x+27}\).

(i) Let

\(y=\frac1{2x-5}.\)

Rearrange to make \(x\) the subject:

\(2x-5=\frac1y.\)

So

\(2x=5+\frac1y, \qquad x=\frac12\left(5+\frac1y\right) =\frac{5y+1}{2y}.\)

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

\(\mathrm f^{-1}(x)=\frac{5x+1}{2x}.\)

(ii) Since the original domain is \(x\gt2.5\), we have \(2x-5\gt0\), so \(\mathrm f(x)\gt0\). Therefore the domain of \(\mathrm f^{-1}\) is

\(x\gt0.\)

(iii) \(\mathrm f^2(x)\) means \(\mathrm f(\mathrm f(x))\). Thus

\(\mathrm f^2(x) =\frac1{2\left(\frac1{2x-5}\right)-5}.\)

Simplify the denominator:

\(2\left(\frac1{2x-5}\right)-5 =\frac2{2x-5}-5 =\frac{2-5(2x-5)}{2x-5}.\)

Therefore

\(\mathrm f^2(x) =\frac{2x-5}{2-5(2x-5)} =\frac{2x-5}{-10x+27}.\)

Answer: (i) \(\mathrm f^{-1}(x)=\dfrac{5x+1}{2x}\). (ii) \(x\gt0\). (iii) \(\mathrm f^2(x)=\dfrac{2x-5}{-10x+27}\).

(i) Let

\(y=\frac1{2x-5}.\)

Rearrange to make \(x\) the subject:

\(2x-5=\frac1y.\)

So

\(2x=5+\frac1y, \qquad x=\frac12\left(5+\frac1y\right) =\frac{5y+1}{2y}.\)

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

\(\mathrm f^{-1}(x)=\frac{5x+1}{2x}.\)

(ii) Since the original domain is \(x\gt2.5\), we have \(2x-5\gt0\), so \(\mathrm f(x)\gt0\). Therefore the domain of \(\mathrm f^{-1}\) is

\(x\gt0.\)

(iii) \(\mathrm f^2(x)\) means \(\mathrm f(\mathrm f(x))\). Thus

\(\mathrm f^2(x) =\frac1{2\left(\frac1{2x-5}\right)-5}.\)

Simplify the denominator:

\(2\left(\frac1{2x-5}\right)-5 =\frac2{2x-5}-5 =\frac{2-5(2x-5)}{2x-5}.\)

Therefore

\(\mathrm f^2(x) =\frac{2x-5}{2-5(2x-5)} =\frac{2x-5}{-10x+27}.\)

Answer: \(2\leq f(x)\leq4\); \(f^{-1}(2.5)=\dfrac{\pi}{6}\); \(x=0,\ \pm\sqrt6\).

For \(0\leq x\leq\frac{\pi}{2}\), we have \(0\leq2x\leq\pi\). Therefore \(\cos2x\) decreases from \(1\) to \(-1\).

So

\(f(x)=3-\cos2x\)

takes values from \(2\) to \(4\). Hence the range is

\(2\leq f(x)\leq4.\)

To find \(f^{-1}(2.5)\), solve

\(3-\cos2x=2.5.\)

Thus

\(\cos2x=0.5.\)

Since \(0\leq2x\leq\pi\),

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

Therefore

\(f^{-1}(2.5)=x=\frac{\pi}{6}.\)

For \(g(x)=3-x^2\),

\(g^2(x)=g(g(x))=3-(3-x^2)^2.\)

Set this equal to \(-6\):

\(3-(3-x^2)^2=-6.\)

So

\((3-x^2)^2=9.\)

Hence

\(3-x^2=3\quad\text{or}\quad 3-x^2=-3.\)

The first equation gives \(x=0\). The second gives \(x^2=6\), so \(x=\pm\sqrt6\).

Therefore

\(\boxed{x=0,\ \pm\sqrt6}.\)

Answer: (i) \(5\leq f(x)\leq6\). (ii) \(f^{-1}(x)=4\sin^{-1}(x-5)\), range \(0\leq f^{-1}(x)\leq2\pi\). (iii) \(x=\pi\).

(i) Since \(0\leq x\leq2\pi\),

\(0\leq \frac{x}{4}\leq \frac{\pi}{2}.\)

On this interval, \(\sin\dfrac{x}{4}\) ranges from \(0\) to \(1\). Therefore

\(5\leq f(x)\leq6.\)

(ii) Let \(y=f(x)\). Then

\(y=5+\sin\frac{x}{4}.\)

So

\(y-5=\sin\frac{x}{4}.\)

Taking inverse sine,

\(\frac{x}{4}=\sin^{-1}(y-5),\)

and hence

\(x=4\sin^{-1}(y-5).\)

Therefore

\(f^{-1}(x)=4\sin^{-1}(x-5).\)

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

\(0\leq f^{-1}(x)\leq2\pi.\)

(iii) Here \(fg(x)\) means \(f(g(x))\). Thus

\(2f(g(x))=11.\)

So

\(f(g(x))=\frac{11}{2}.\)

Since \(g(x)=x-\dfrac{\pi}{3}\),

\(f(g(x))=5+\sin\left(\frac{x-\frac{\pi}{3}}{4}\right).\)

Therefore

\(5+\sin\left(\frac{x-\frac{\pi}{3}}{4}\right)=\frac{11}{2}.\)

Hence

\(\sin\left(\frac{x-\frac{\pi}{3}}{4}\right)=\frac12.\)

The input of \(f\) must lie in \(0\leq g(x)\leq2\pi\), so

\(0\leq x-\frac{\pi}{3}\leq2\pi.\)

Thus

\(0\leq \frac{x-\frac{\pi}{3}}{4}\leq\frac{\pi}{2}.\)

On this interval, the only solution to \(\sin u=\dfrac12\) is \(u=\dfrac{\pi}{6}\). Therefore

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

So

\(x=\frac{4\pi}{6}+\frac{\pi}{3} =\frac{2\pi}{3}+\frac{\pi}{3} =\pi.\)

Answer: range \(f(x)\gt 5\); \(f^{-1}(x)=-\dfrac14\ln\!\left(\dfrac{x-5}{3}\right)\), domain \(x\gt 5\); \(x=\sqrt{e^2-5}\approx1.55\).

Since \(e^{-4x}\gt 0\) for all real \(x\),

\(3e^{-4x}+5\gt 5.\)

Therefore the range is

\(f(x)\gt 5.\)

To find the inverse, let

\(y=3e^{-4x}+5.\)

Then

\(y-5=3e^{-4x}.\)

So

\(\frac{y-5}{3}=e^{-4x}.\)

Take natural logarithms:

\(\ln\left(\frac{y-5}{3}\right)=-4x.\)

Hence

\(x=-\frac14\ln\left(\frac{y-5}{3}\right).\)

Therefore

\(f^{-1}(x)=-\frac14\ln\left(\frac{x-5}{3}\right).\)

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

\(x\gt 5.\)

For part (b),

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

So

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

Therefore

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

Thus

\(x=\sqrt{e^2-5}\approx1.55.\)

Answer: range of \(f\): \(f(x)\geq3\); \(x=0,\dfrac12\); \(h^{-1}(x)=\dfrac{4x+1}{x-2}\), range \(x\neq4\); \(h^2(x)=\dfrac{5x-2}{17-2x}\).

Since \(f(x)=x^2+3\) and \(x^2\geq0\),

\(f(x)\geq3.\)

So the range of \(f\) is \(f(x)\geq3\).

Now \(fg(x)\) means \(f(g(x))\):

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

Set this equal to \(4\):

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

Then

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

So \(4x-1=1\) or \(4x-1=-1\), giving

\(\boxed{x=\dfrac12\quad\text{or}\quad x=0}.\)

For the inverse of \(h\), write

\(y=\dfrac{2x+1}{x-4}.\)

Multiply by \(x-4\):

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

So

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

Collect the \(x\)-terms:

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

Therefore

\(x=\dfrac{4y+1}{y-2}.\)

Hence

\(h^{-1}(x)=\dfrac{4x+1}{x-2}.\)

The range of \(h^{-1}\) is the domain of \(h\), so it is all real values except \(4\).

Finally,

\(h^2(x)=h(h(x)).\)

Substitute \(h(x)=\dfrac{2x+1}{x-4}\) into \(h\):

\(h^2(x)=\dfrac{2\left(\dfrac{2x+1}{x-4}\right)+1}{\left(\dfrac{2x+1}{x-4}\right)-4}.\)

Simplifying the numerator:

\(\dfrac{4x+2+x-4}{x-4}=\dfrac{5x-2}{x-4}.\)

Simplifying the denominator:

\(\dfrac{2x+1-4x+16}{x-4}=\dfrac{17-2x}{x-4}.\)

Therefore

\(h^2(x)=\dfrac{5x-2}{17-2x}.\)

Answer: range of \(f\): all real numbers; range of \(g\): \(y\gt 3\); \(f^{-1}g(4)=e^{35}\); \(g^{-1}(x)=\sqrt{\dfrac{x-3}{2}}\), domain \(x\gt 3\).

Since \(f(x)=\ln x\) for \(x\gt 0\), the range of \(f\) is all real numbers.

For \(g(x)=2x^2+3\), with \(x\gt 0\), we have \(2x^2\gt 0\). Therefore the range of \(g\) is

\(y\gt 3.\)

Now

\(g(4)=2(4)^2+3=35.\)

Since \(f(x)=\ln x\), the inverse is \(f^{-1}(x)=e^x\). Therefore

\(f^{-1}g(4)=f^{-1}(35)=\boxed{e^{35}}.\)

To find \(g^{-1}\), write

\(y=2x^2+3.\)

Then

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

Since the domain of \(g\) is \(x\gt 0\), take the positive square root:

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

Hence

\(g^{-1}(x)=\sqrt{\dfrac{x-3}{2}}.\)

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

\(x\gt 3.\)

Answer: \(f^2(-3)=17\); \(x=1\) or \(x=-5\).

First,

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

Therefore

\(f^2(-3)=f(f(-3))=f(2)=(2+2)^2+1=17.\)

To find the inverse of \(g\), let

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

Then

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

So

\(2xy-y=x-2.\)

Collect the \(x\)-terms:

\(2xy-x=y-2.\)

Factorise:

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

Therefore

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

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

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

Now

\(gf(x)=g((x+2)^2+1) = \frac{(x+2)^2+1-2}{2((x+2)^2+1)-1}.\)

So

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

Set this equal to \(\dfrac8{19}\):

\(\frac{(x+2)^2-1}{2(x+2)^2+1}=\frac8{19}.\)

Cross-multiply:

\(19((x+2)^2-1)=8(2(x+2)^2+1).\)

Let \(u=(x+2)^2\). Then

\(19(u-1)=8(2u+1).\)

So

\(19u-19=16u+8,\qquad 3u=27,\qquad u=9.\)

Hence

\((x+2)^2=9.\)

Thus \(x+2=\pm3\), giving

\(\boxed{x=1\ \text{or}\ x=-5}.\)

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

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