What is the extreme f(x)=(3t^2),(0<= t<= 15) ?

The extreme f(x)=(3t^2),(0<= t<= 15) is Minimum(0,0),Maximum(15,675)

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extreme f(x)=(3t^2),(0<= t<= 15)

extreme points

f (x) = (3 t^{2}), (0 \leq t \leq 15)

Minimum (0, 0), Maximum (15, 675)

Solution steps

Find the critical points:

t = 0, t = 15

Domain of

3 t^{2}, 0 \leq t \leq 15 : 0 \leq t \leq 15

Find intervals:

Increasing

: 0 < t < 15

Plug the extreme point

x = 0

into

3 t^{2} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 15

into

3 t^{2} \Rightarrow y = 675

Maximum (15, 675)

Minimum (0, 0), Maximum (15, 675)

f (x, y) = x^{3} + y^{2} - 6 x^{2} + y - 1

x^{2} - x - 12

f (x) = x^{2} y + y^{3} - 75 y

f (x, y) = x^{2} + x y + y^{2} - 4 y + 5

y = x^{3} - 3 x^{2} + 1

What is the extreme f(x)=(3t^2),(0<= t<= 15) ?
The extreme f(x)=(3t^2),(0<= t<= 15) is Minimum(0,0),Maximum(15,675)