What is the extreme f(x)=sqrt(36-t^2) ?

The extreme f(x)=sqrt(36-t^2) is Minimum(-6,0),Maximum(0,6),Minimum(6,0)

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extreme f(x)=sqrt(36-t^2)

extreme points

f (x) = 36 - t^{2}

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

Solution steps

f^{'} (t) = - \frac{t}{36 - t ^{2}}

Find intervals:

Increasing

: - 6 < t < 0,

Decreasing

: 0 < t < 6

Plug

t = - 6

into

36 - t^{2} : 0

Minimum (- 6, 0)

Plug

t = 0

into

36 - t^{2} : 6

Maximum (0, 6)

Plug

t = 6

into

36 - t^{2} : 0

Minimum (6, 0)

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

- x^{3} + 2 x^{2} + 4 x + 7

f (x) = 2 x^{3} - 39 x^{2} + 180 x + 1

y = 6 x - 12

1353.0186700000000 \dots

f (x) = \frac{1}{8} (x + 4)^{2} (6 - x)

What is the extreme f(x)=sqrt(36-t^2) ?
The extreme f(x)=sqrt(36-t^2) is Minimum(-6,0),Maximum(0,6),Minimum(6,0)