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

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

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

extreme points

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

Maximum (- 1, \frac{1}{3}), Minimum (0, 0), Maximum (1, \frac{1}{3})

Solution steps

f^{'} (t) = 4 t^{3} - 4 t^{5}

Find intervals:

Increasing

: - \infty < t < - 1,

Decreasing

: - 1 < t < 0,

Increasing

: 0 < t < 1,

Decreasing

: 1 < t < \infty

Plug

t = - 1

into

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

Maximum (- 1, \frac{1}{3})

Plug

t = 0

into

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

Minimum (0, 0)

Plug

t = 1

into

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

Maximum (1, \frac{1}{3})

Maximum (- 1, \frac{1}{3}), Minimum (0, 0), Maximum (1, \frac{1}{3})

f (x) = 7 sin (x)

\frac{x ^{3} - x}{x ^{2} - 4}

f (x) = 3 + 1

y = x^{2} - x - 20

\frac{2 x - 1}{x ^{2} - 1}

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