Q. The function $\mathrm{f}:[\mathrm{a}, \infty) \rightarrow \mathrm{R}$ where $\mathrm{R}$ denotes the range corresponding to the given domain, with rule $f(x)=2 x^{3}-3 x^{2}+6$, will have an inverse provided
Application of Derivatives
Solution:
$\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-3 \mathrm{x}^{2}+6$
$f^{\prime}(x)=6 x^{2}-6 x=6\left(x^{2}-x\right)=0 $ gives $ x=0$ or $x=1$
for inverse to exist function must be one one onto
hence Domain is $[1, \infty)$
Hence $a \geq 1$
