What is the extreme f(x)=x^3-9x^2+15x-6 ?

The extreme f(x)=x^3-9x^2+15x-6 is Maximum(1,1),Minimum(5,-31)

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extreme f(x)=x^3-9x^2+15x-6

extreme points

f (x) = x^{3} - 9 x^{2} + 15 x - 6

Maximum (1, 1), Minimum (5, - 31)

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 - 6 : 1

Maximum (1, 1)

Plug

x = 5

into

x^{3} - 9 x^{2} + 15 x - 6 : - 31

Minimum (5, - 31)

Maximum (1, 1), Minimum (5, - 31)

(x^{2} - 7)^{3}

f (x) = \frac{5}{4 x - 8}

\frac{x ^{2} - 16}{3 x ^{2}}

f (x) = t 4 - t

x^{2} - 2

What is the extreme f(x)=x^3-9x^2+15x-6 ?
The extreme f(x)=x^3-9x^2+15x-6 is Maximum(1,1),Minimum(5,-31)