Answer: (a) \(f(x)\gt -4\); (b) \(\displaystyle f^{-1}(x)=\frac13\ln(x+4)\); (c) intercepts for \(f\): \(\left(\frac13\ln4,0\right)\), \((0,-3)\); intercepts for \(f^{-1}\): \((-3,0)\), \(\left(0,\frac13\ln4\right)\).
Answer: (a) \(f(x)\gt -4\); (b) \(\displaystyle f^{-1}(x)=\frac13\ln(x+4)\); (c) intercepts for \(f\): \(\left(\frac13\ln4,0\right)\), \((0,-3)\); intercepts for \(f^{-1}\): \((-3,0)\), \(\left(0,\frac13\ln4\right)\).
(a) Since \(e^{3x}\gt 0\) for all real \(x\),
\(e^{3x}-4\gt -4.\)
So the range is
\(f(x)\gt -4.\)
(b) Let
\(y=e^{3x}-4.\)
Then
\(y+4=e^{3x}.\)
Taking natural logarithms,
\(\ln(y+4)=3x.\)
So
\(x=\frac13\ln(y+4).\)
Therefore
\(f^{-1}(x)=\frac13\ln(x+4).\)
(c) For \(y=f(x)\), the \(y\)-intercept is
\(f(0)=e^0-4=-3.\)
The \(x\)-intercept satisfies
\(e^{3x}-4=0.\)
So
\(e^{3x}=4,\)
and hence
\(x=\frac13\ln4.\)
Thus \(y=f(x)\) has intercepts
\((0,-3)\quad\text{and}\quad\left(\frac13\ln4,0\right).\)
The graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\). Its intercepts are therefore
\((-3,0)\quad\text{and}\quad\left(0,\frac13\ln4\right).\)
