What is the extreme f(x)=(x^3)/3-4x^2+7x+8 ?

The extreme f(x)=(x^3)/3-4x^2+7x+8 is Maximum(1, 34/3),Minimum(7,-74/3)

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extreme f(x)=(x^3)/3-4x^2+7x+8

extreme points

f (x) = \frac{x ^{3}}{3} - 4 x^{2} + 7 x + 8

Maximum (1, \frac{34}{3}), Minimum (7, - \frac{74}{3})

Solution steps

f^{'} (x) = x^{2} - 8 x + 7

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = 1

into

\frac{x ^{3}}{3} - 4 x^{2} + 7 x + 8 : \frac{34}{3}

Maximum (1, \frac{34}{3})

Plug

x = 7

into

\frac{x ^{3}}{3} - 4 x^{2} + 7 x + 8 : - \frac{74}{3}

Minimum (7, - \frac{74}{3})

Maximum (1, \frac{34}{3}), Minimum (7, - \frac{74}{3})

What is the extreme f(x)=(x^3)/3-4x^2+7x+8 ?
The extreme f(x)=(x^3)/3-4x^2+7x+8 is Maximum(1, 34/3),Minimum(7,-74/3)