The value of `c` in Rolles theorem when `f(x)=2x^3-5x^2-4x+3` , is `x in [1//3,\ 3]`

Answer

`C` is `2`

Explanation

Given:

`f(x)=2 x^{3}-5 x^{2}-4 x+3`


Differentiating the given function w.r.t `x`, we will get

$$
\begin{aligned}
&f^{\prime}(x)=6 x^{2}-10 x-4 \\
&\Rightarrow f^{\prime}(c)=6 c^{2}-10 c-4 \\
&\therefore f^{\prime}(c)=0 \\
...