# How do I find the numbers c that satisfy the Mean Value Theorem for f(x)=3x^2+2x+5 on the interval [-1,1] ?

May 2, 2018

$c = 0$

#### Explanation:

We seek to verify the Mean Value Theorem for the function

$f \left(x\right) = 3 {x}^{2} + 2 x + 5$ on the interval $\left[- 1 , 1\right]$

The Mean Value Theorem, tells us that if $f \left(x\right)$ is differentiable on a interval $\left[a , b\right]$ then $\exists \setminus c \in \left[a , b\right]$ st:

$f ' \left(c\right) = \frac{f \left(b\right) - f \left(a\right)}{b - a}$

So, Differentiating wrt $x$ we have:

$f ' \left(x\right) = 6 x + 2$

And we seek a value $c \in \left[- 1 , 1\right]$ st: $f ' \left(c\right) = \frac{f \left(1\right) - f \left(- 1\right)}{1 - \left(- 1\right)}$

$\therefore 6 c + 2 = \frac{\left(3 + 2 + 5\right) - \left(3 - 2 + 5\right)}{2}$

$\therefore 6 c + 2 = \frac{4}{2}$

$\therefore 6 c + 2 = 2$

$\therefore 6 c = 0$

$\therefore c = 0$