Answer: (a) \(-1.5\); (b) all real values; (c) \(\displaystyle f^{-1}(x)=\frac{e^x-12}{8}\), range \(f^{-1}(x)\gt -1.5\); (d) intercepts \(\left(-\frac{11}{8},0\right)\), \((0,\ln12)\), \(\left(\ln12,0\right)\), \(\left(0,-\frac{11}{8}\right)\).
Answer: (a) \(-1.5\); (b) all real values; (c) \(\displaystyle f^{-1}(x)=\frac{e^x-12}{8}\), range \(f^{-1}(x)\gt -1.5\); (d) intercepts \(\left(-\frac{11}{8},0\right)\), \((0,\ln12)\), \(\left(\ln12,0\right)\), \(\left(0,-\frac{11}{8}\right)\).
(a) The logarithm requires
\(2x+3\gt 0.\)
So
\(x\gt -\frac32.\)
Hence the least possible value of \(a\) is
\(-\frac32=-1.5.\)
(b) Using this domain, \(2x+3\) can take all positive values. Therefore \(\ln(2x+3)+\ln4\) can take all real values. The range of \(f\) is
\(f(x)\in\mathbb R.\)
(c) Write
\(y=\ln(2x+3)+\ln4.\)
Use the logarithm law:
\(y=\ln(4(2x+3))=\ln(8x+12).\)
Then
\(e^y=8x+12.\)
So
\(x=\frac{e^y-12}{8}.\)
Therefore
\(f^{-1}(x)=\frac{e^x-12}{8}.\)
The range of \(f^{-1}\) is the domain of \(f\), so
\(f^{-1}(x)\gt -\frac32.\)
(d) For \(y=f(x)\), the \(x\)-intercept is found from
\(\ln(8x+12)=0.\)
So
\(8x+12=1,\)
giving
\(x=-\frac{11}{8}.\)
The \(y\)-intercept is
\(f(0)=\ln12.\)
So \(y=f(x)\) has intercepts
\(\left(-\frac{11}{8},0\right)\quad\text{and}\quad(0,\ln12).\)
The graph of \(y=f^{-1}(x)\) is the reflection of \(y=f(x)\) in the line \(y=x\). Its intercepts are therefore
\((\ln12,0)\quad\text{and}\quad\left(0,-\frac{11}{8}\right).\)
