What is the extreme f(t)=4t^3-36t^2+10,t>= 0 ?

The extreme f(t)=4t^3-36t^2+10,t>= 0 is Maximum(0,10),Minimum(6,-422)

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extreme f(t)=4t^3-36t^2+10,t>= 0

extreme points

f (t) = 4 t^{3} - 36 t^{2} + 10, t \geq 0

Maximum (0, 10), Minimum (6, - 422)

Solution steps

Find the critical points:

t = 0, t = 6

Domain of

4 t^{3} - 36 t^{2} + 10, t \geq 0 : t \geq 0

Find intervals:

Decreasing

: 0 < t < 6,

Increasing

: 6 < t < \infty

Plug the extreme point

x = 0

into

4 t^{3} - 36 t^{2} + 10 \Rightarrow y = 10

Maximum (0, 10)

Plug the extreme point

x = 6

into

4 t^{3} - 36 t^{2} + 10 \Rightarrow y = - 422

Minimum (6, - 422)

Maximum (0, 10), Minimum (6, - 422)

f (x) = 3 x^{4} + 20 x^{3} - 36 x^{2} - 4

2 x^{3} - 30 x^{2}, - 1 \leq x \leq 11

f (x) = x (1 - x)^{\frac{2}{5}}

f (x) = x^{2} - 2 x + 2, - 2 \leq x \leq 2

\frac{6}{- 3 x + 2}

What is the extreme f(t)=4t^3-36t^2+10,t>= 0 ?
The extreme f(t)=4t^3-36t^2+10,t>= 0 is Maximum(0,10),Minimum(6,-422)