Consider the function $f (x) = 4 x^{3} - 4 x$ $f(x) = 4 x^3 − 4 x$ on the interval [ −3 , 3 ], how do you find the average or mean slope of the function on this interval?

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Answer 1 · 2017-10-26T13:51:30

See the explanation below

The average or mean slope of a function $f (x)$ $f(x)$ on the interval $[a, b]$ $[a,b]$ is

$= \frac{f (b) - f (a)}{b - a}$ $=(f(b)-f(a))/(b-a)$

Here,

$f (x) = 4 x^{3} - 4 x = 4 x (x^{2} - 1)$ $f(x)=4x^3-4x=4x(x^2-1)$

and the interval is $= [- 3, 3]$ $=[-3,3]$

$f (3) = 12 (8) = 96$ $f(3)=12(8)=96$

$f (- 3) = - 12 \cdot 8 = - 96$ $f(-3)=-12*8=-96$

Therefore,

The mean slope is

$= \frac{f (3) - f (3)}{3 - (- 3)} = \frac{96 + 96}{6} = 32$ $=(f(3)-f(3))/(3-(-3))=(96+96)/(6)=32$

Then,

There exists $c \in (- 3, 3)$ $c in (-3,3)$ such that

$f'(c)=32$

$f'(x)=12x^2-4=4(3x^2-1)$

Therefore,

$f'(c)=4(3c^2-1)=32$

$3c^2-1=8$

$c^2=9/3=3$

$c=+-sqrt3$

So,

$+-sqrt3 in (-3,3)$

Consider the function f(x) = 4 x^3 − 4 xf(x)=4x3−4xf(x) = 4 x^3 − 4 x on the interval [ −3 , 3 ], how do you find the average or mean slope of the function on this interval?