For the function

The given function is $f(x)=8 x^{2}-7 x+5$.

f(x) is a polynomial function.

We know that a polynomial function is everywhere continuous and differentiable. So, $f(x)$ is continuous on $[-6,6]$ and differentiable on $(-6,6)$. Thus, both the conditions of Lagrange's mean value theorem are satisfied.

So, there must exist at least one real number $c \in(-6,6)$ such that

$f^{\prime}(c)=\frac{f(6)-f(-6)}{6-(-6)}$

Now, $f(x)=8 x^{2}-7 x+5$

$\Rightarrow f^{\prime}(x)=16 x-7$

$\therefore f^{\prime}(c)=\frac{f(6)-f(-6)}{6-(-6)}$

$\Rightarrow 16 c-7=\frac{\left[8 \times(6)^{2}-7 \times 6+5\right]-\left[8 \times(-6)^{2}-7 \times(-6)+5\right]}{12}$

$\Rightarrow 16 c-7=\frac{-84}{12}=-7$

$\Rightarrow 16 c=0$

$\Rightarrow c=0$

Thus, $c=0 \in(-6,6)$ such that $f^{\prime}(c)=\frac{f(6)-f(-6)}{6-(-6)}$.

Hence, the value of c is 0.

For the function $f(x)=8 x^{2}-7 x+5, x \in[-6,6]$, the value of $c$ for the Lagrange's mean value theorem is __0___.