Question

Find the inverse of the following function:

$f\left(x\right)={\log }_e(x^2+3x+1),x\in [1,3]$

• A. $f^{-1}\left(x\right)=\frac{3+\sqrt{5+4e^x}}{2}$
• B. $f^{-1}\left(x\right)=\frac{-3+\sqrt{5+4e^x}}{2}$
• C. $f^{-1}\left(x\right)=\frac{-3-\sqrt{5+4e^x}}{2}$
• D. $f^{-1}\left(x\right)=\frac{3-\sqrt{5-4e^x}}{2}$

Answer 1

Answer: (B)

$f^{-1}\left(x\right)=\frac{-3+\sqrt{5+4e^x}}{2}$

$y=ln(x^2+3x+1)$ domain is $xϵ[1,3]$
Hence $f\left(x\right)=y$ is invertible in this domain.
Therefore
$y=ln(x^2+3x+1)$
$e^y=x^2+3x+1$
$e^y=(x+\frac{3}{2}{)}^2-\frac{9}{4}+1$
$e^4=(x+\frac{3}{2}{)}^2-\frac{5}{4}$
$\frac{5+4e^y}{4}=(x+\frac{3}{2}{)}^2$
$\frac{\pm \sqrt{5+4e^y}}{2}=x+\frac{3}{2}$
$x=\frac{-3\pm \sqrt{5+4e^y}}{2}$
Replacing x and y gives us
$f^{-1}\left(x\right)=\frac{-3\pm \sqrt{5+4e^x}}{2}$
Now if
$f^{-1}\left(x\right)=\frac{-3-\sqrt{5+4e^x}}{2}$
$f^{-1}\left(x\right)<0$ for all $xϵ[1,3]$.
But we know that the range of the above function is strictly positive.
Hence
$f^{-1}\left(x\right)=\frac{-3+\sqrt{5+4e^x}}{2}$