$x^{3} + 24 x - 16$ $x^3+24x-16$ [0,4] verify mean value theorem?

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Answer 1 · 2018-05-02T13:54:59

We seek to verify the Mean Value Theorem for the function

$f (x) = x^{3} + 24 x - 16$ $f(x) = x^3+24x-16$ on the interval $[0, 4]$ $[0,4]$

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

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

So, Differentiating wrt $x$ we have:

$f'(x) = 3x^2 + 24$

And we seek a value $c in [0,4]$ st: $f'(c) = (f(4)-f(0))/(4-0)$

$:. 3c^2 + 24 = ((64+96-16)-(0+0-16))/(4-0)$

$:. 3c^2 + 24 = 160/4$

$:. 3c^2 + 24 = 40$

$:. 3c^2 = 16$

$:. c^2 = 16/3$

$:. c = +- (4sqrt(3))/3$

And we require that $c in [0,4]$ , so we choose $c=(4sqrt(3))/3$

x^3+24x-16x3+24x−16x^3+24x-16 [0,4] verify mean value theorem?