Question

For the function f(x) = 8x² - 7x + 5, x ∈ [-6, 6], the value of c for the lagrange's mean value theorem is __________________.

Answer 1

The given function is $f\left(x\right)=8x^2-7x+5$.

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 ∈ (−6, 6) such that

$f\text{'}\left(c\right)=\frac{f\left(6\right)-f\left(-6\right)}{6-\left(-6\right)}$

Now, $f\left(x\right)=8x^2-7x+5$

$\Rightarrow f\text{'}\left(x\right)=16x-7$

$\therefore f\text{'}\left(c\right)=\frac{f\left(6\right)-f\left(-6\right)}{6-\left(-6\right)}$

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

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

$\Rightarrow 16c=0$

$\Rightarrow c=0$

Thus, c = 0 ∈ (−6, 6) such that $f\text{'}\left(c\right)=\frac{f\left(6\right)-f\left(-6\right)}{6-\left(-6\right)}$.

Hence, the value of c is 0.


For the function f(x) = 8x² − 7x + 5, x ∈ [−6, 6], the value of c for the Lagrange's mean value theorem is ___0___.