What is the extreme f(x)=x^3-9x^2+5 ?

The extreme f(x)=x^3-9x^2+5 is Maximum(0,5),Minimum(6,-103)

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extreme f(x)=x^3-9x^2+5

extreme points

f (x) = x^{3} - 9 x^{2} + 5

Maximum (0, 5), Minimum (6, - 103)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

x^{3} - 9 x^{2} + 5 : 5

Maximum (0, 5)

Plug

x = 6

into

x^{3} - 9 x^{2} + 5 : - 103

Minimum (6, - 103)

Maximum (0, 5), Minimum (6, - 103)

\frac{5 x + 1}{7 x + 9}

f (x) = 2 x + 5 y = 20

f (x) = 6 - 5 x^{3}

f (t) = sin (2 t - \frac{π}{3}) - 4

f (x) = \frac{x ^{2} - 5 x - 3}{2 x + 1}

What is the extreme f(x)=x^3-9x^2+5 ?
The extreme f(x)=x^3-9x^2+5 is Maximum(0,5),Minimum(6,-103)