What is the extreme y=x^3-6x^2+15 ?

The extreme y=x^3-6x^2+15 is Maximum(0,15),Minimum(4,-17)

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extreme y=x^3-6x^2+15

extreme points

y = x^{3} - 6 x^{2} + 15

Maximum (0, 15), Minimum (4, - 17)

Solution steps

f^{'} (x) = 3 x^{2} - 12 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

x^{3} - 6 x^{2} + 15 : 15

Maximum (0, 15)

Plug

x = 4

into

x^{3} - 6 x^{2} + 15 : - 17

Minimum (4, - 17)

Maximum (0, 15), Minimum (4, - 17)

f (x) = (\frac{x ^{2} - 1}{( 5 - x ^{2} ) ^{0.5}})

f (x) = 2 x^{3} + 9 x^{2} - 60 x + 1

f (x) = 2 x^{3} + 9 x^{2} - 60 x + 8

y = x^{\frac{4}{7}} (x^{2} - 2)

f (θ) = 2 θ - 3 sin (θ), 0 \leq θ \leq π

What is the extreme y=x^3-6x^2+15 ?
The extreme y=x^3-6x^2+15 is Maximum(0,15),Minimum(4,-17)