What is the extreme f(x)=(t^2-4)^3 ?

The extreme f(x)=(t^2-4)^3 is Saddle(-2,0),Minimum(0,-64),Saddle(2,0)

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extreme f(x)=(t^2-4)^3

extreme points

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

Saddle (- 2, 0), Minimum (0, - 64), Saddle (2, 0)

Solution steps

f^{'} (t) = 6 t (t^{2} - 4)^{2}

Find intervals:

Decreasing

: - \infty < t < - 2,

Decreasing

: - 2 < t < 0,

Increasing

: 0 < t < 2,

Increasing

: 2 < t < \infty

Plug

t = - 2

into

(t^{2} - 4)^{3} : 0

Saddle (- 2, 0)

Plug

t = 0

into

(t^{2} - 4)^{3} : - 64

Minimum (0, - 64)

Plug

t = 2

into

(t^{2} - 4)^{3} : 0

Saddle (2, 0)

Saddle (- 2, 0), Minimum (0, - 64), Saddle (2, 0)

\frac{x}{x ^{2} + 16}, 0 \leq x \leq 8

f (x) = \frac{( x ^{2} - 5 x + 2 )}{x - 5}

(8 - x) (x + 1)^{2}

f (x) = x^{3} + 11 x + 10

f (x) = 3 sin (x), 0 \leq x \leq 2 π

What is the extreme f(x)=(t^2-4)^3 ?
The extreme f(x)=(t^2-4)^3 is Saddle(-2,0),Minimum(0,-64),Saddle(2,0)