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

The extreme f(x)=t^4-2t^2+2 is Minimum(-1,1),Maximum(0,2),Minimum(1,1)

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

extreme points

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

Minimum (- 1, 1), Maximum (0, 2), Minimum (1, 1)

Solution steps

f^{'} (t) = 4 t^{3} - 4 t

Find intervals:

Decreasing

: - \infty < t < - 1,

Increasing

: - 1 < t < 0,

Decreasing

: 0 < t < 1,

Increasing

: 1 < t < \infty

Plug

t = - 1

into

t^{4} - 2 t^{2} + 2 : 1

Minimum (- 1, 1)

Plug

t = 0

into

t^{4} - 2 t^{2} + 2 : 2

Maximum (0, 2)

Plug

t = 1

into

t^{4} - 2 t^{2} + 2 : 1

Minimum (1, 1)

Minimum (- 1, 1), Maximum (0, 2), Minimum (1, 1)

f (x) = (x + 2)^{3} (x - 3)^{2}

y = (x - 1)^{2} (x + 1)

f (x) = (x + 2)^{3} (x - 3)^{3}

f (x) = 5 x^{2} - 20 x + 5, 0 \leq x \leq 3

f (x) = x^{\frac{2}{3}} (\frac{1}{8} x^{2} + \frac{3}{5} x - 9)

What is the extreme f(x)=t^4-2t^2+2 ?
The extreme f(x)=t^4-2t^2+2 is Minimum(-1,1),Maximum(0,2),Minimum(1,1)