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

\(x=64\).