Question

Let, $f:A\to B$ be an invertible function. If $f\left(x\right)=2x^3+3x^2+x-1$, then $f^{\prime -1}\left(5\right)$=

• A. $\frac{1}{13}$
• B. $1$
• C. $6$
• D. can not be determined

Answer 1

The correct option is A $\frac{1}{13}$

Given $f$ be an invertible function and $f\left(x\right)=2x^3+3x^2+x-1$.

Then $f^{\prime -1}\left(5\right)=\frac{1}{f^{\prime }\left(x\right)}\forall x\in f\left(A\right)$ such that $f\left(x\right)=5$.

Now, $f\left(x\right)=5$ given only $x=1$ a real root, remaining are imaginary.

Now, $f^{\prime }\left(x\right)=6(x^2+x)+1$.

$\therefore$$f^{\prime -1}\left(5\right)=\frac{1}{f^{\prime }\left(1\right)}=\frac{1}{13}$