Question

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

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Solution

## The given function is $f\left(x\right)=8{x}^{2}-7x+5$. 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, $f\left(x\right)=8{x}^{2}-7x+5$ $⇒f\text{'}\left(x\right)=16x-7$ $\therefore f\text{'}\left(c\right)=\frac{f\left(6\right)-f\left(-6\right)}{6-\left(-6\right)}$ $⇒16c-7=\frac{\left[8×{\left(6\right)}^{2}-7×6+5\right]-\left[8×{\left(-6\right)}^{2}-7×\left(-6\right)+5\right]}{12}$ $⇒16c-7=\frac{-84}{12}=-7$ $⇒16c=0$ $⇒c=0$ 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___.

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