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

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Answer 1 · 2018-05-02T14:09:51

$c=0$

We seek to verify the Mean Value Theorem for the function

$f(x) = 3x^2+2x+5$ on the interval $[-1,1]$

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

$f'(c) = (f(b)-f(a))/(b-a)$

So, Differentiating wrt $x$ we have:

$f'(x) = 6x+2$

And we seek a value $c in [-1,1]$ st: $f'(c) = (f(1)-f(-1))/(1-(-1))$

$:. 6c+2 = ((3+2+5)-(3-2+5))/(2)$

$:. 6c + 2 = 4/2$

$:. 6c + 2 = 2$

$:. 6c = 0$

$:. c=0$

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