Question

Prove that the function $f[0,\infty )\to R$ given by $f\left(x\right)=9x^2+6x-5$ is not the codomain of the function f to make it invertible, and hence find $f^{\prime }$

Answer 1

$f\left(x\right)=9x^2+6x-5$

The min value of $f\left(x\right)$ is at $x=-b/a=\frac{-6}{9}=-\frac{2}{3}$

$f(-2/3)=9\times \frac{4}{9}+6\times \frac{2}{3}-54+4-5=3$

The function has no value less than $3$ but $R$ contains values less than $3$

So $f$ is not invertible.

$f^{\prime }\left(x\right)=\frac{d}{dx}(9x^2+6x-5)=18x+6$