What is the extreme f(x)=(t^2-1)^3 ?

The extreme f(x)=(t^2-1)^3 is Saddle(-1,0),Minimum(0,-1),Saddle(1,0)

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extreme f(x)=(t^2-1)^3

extreme points

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

Saddle (- 1, 0), Minimum (0, - 1), Saddle (1, 0)

Solution steps

f^{'} (t) = 6 t (t^{2} - 1)^{2}

Find intervals:

Decreasing

: - \infty < t < - 1,

Decreasing

: - 1 < t < 0,

Increasing

: 0 < t < 1,

Increasing

: 1 < t < \infty

Plug

t = - 1

into

(t^{2} - 1)^{3} : 0

Saddle (- 1, 0)

Plug

t = 0

into

(t^{2} - 1)^{3} : - 1

Minimum (0, - 1)

Plug

t = 1

into

(t^{2} - 1)^{3} : 0

Saddle (1, 0)

Saddle (- 1, 0), Minimum (0, - 1), Saddle (1, 0)

88 - \frac{1}{2} x - \frac{3200}{x}

f (x) = 2 x - 4 sin (x)

f (m, a) = m \cdot a

f (x, y) = \frac{9}{6 - 4 x ^{2} - 9 y ^{2}}

What is the extreme f(x)=(t^2-1)^3 ?
The extreme f(x)=(t^2-1)^3 is Saddle(-1,0),Minimum(0,-1),Saddle(1,0)