What is the extreme f(x)= 3/2 t^4-t^6 ?

The extreme f(x)= 3/2 t^4-t^6 is Maximum(-1, 1/2),Minimum(0,0),Maximum(1, 1/2)

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extreme f(x)= 3/2 t^4-t^6

extreme points

f (x) = \frac{3}{2} t^{4} - t^{6}

Maximum (- 1, \frac{1}{2}), Minimum (0, 0), Maximum (1, \frac{1}{2})

Solution steps

f^{'} (t) = 6 t^{3} - 6 t^{5}

Find intervals:

Increasing

: - \infty < t < - 1,

Decreasing

: - 1 < t < 0,

Increasing

: 0 < t < 1,

Decreasing

: 1 < t < \infty

Plug

t = - 1

into

\frac{3}{2} t^{4} - t^{6} : \frac{1}{2}

Maximum (- 1, \frac{1}{2})

Plug

t = 0

into

\frac{3}{2} t^{4} - t^{6} : 0

Minimum (0, 0)

Plug

t = 1

into

\frac{3}{2} t^{4} - t^{6} : \frac{1}{2}

Maximum (1, \frac{1}{2})

Maximum (- 1, \frac{1}{2}), Minimum (0, 0), Maximum (1, \frac{1}{2})

f (x, y) = - x^{2} - y^{3} + 3 y^{2} - 2 x + 45 y + 1

y = - 12 x^{5} - 45 x^{4} + 20 x^{3} - 1

f (x) = x^{3} - 2 x - 4 x + 4 (0.3)

x^{2} + x^{- 2}

(x - 3)^{2} (x - 2)

What is the extreme f(x)= 3/2 t^4-t^6 ?
The extreme f(x)= 3/2 t^4-t^6 is Maximum(-1, 1/2),Minimum(0,0),Maximum(1, 1/2)