What is the extreme f(x)=x^3-9x^2+15x-4 ?

The extreme f(x)=x^3-9x^2+15x-4 is Maximum(1,3),Minimum(5,-29)

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extreme f(x)=x^3-9x^2+15x-4

extreme points

f (x) = x^{3} - 9 x^{2} + 15 x - 4

Maximum (1, 3), Minimum (5, - 29)

Solution steps

f^{'} (x) = 3 x^{2} - 18 x + 15

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = 1

into

x^{3} - 9 x^{2} + 15 x - 4 : 3

Maximum (1, 3)

Plug

x = 5

into

x^{3} - 9 x^{2} + 15 x - 4 : - 29

Minimum (5, - 29)

Maximum (1, 3), Minimum (5, - 29)

P (x, y) = 3 x^{2} y + 4 x y^{2} - y^{- 4}

f (x) = (x + 10) (\frac{60}{x} + 6)

f (x) = \frac{1}{2} x^{4} - 4 x^{2} + 5

f (x) = \frac{1}{2} x^{4} - 4 x^{2} + 3

f (x) = - 3 x^{2} + 180 x + 190, 0 \leq x \leq 40

What is the extreme f(x)=x^3-9x^2+15x-4 ?
The extreme f(x)=x^3-9x^2+15x-4 is Maximum(1,3),Minimum(5,-29)