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

The extreme f(x)=x^3-8x^2+2 is Maximum(0,2),Minimum(16/3 ,-1994/27)

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

extreme points

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

Maximum (0, 2), Minimum (\frac{16}{3}, - \frac{1994}{27})

Solution steps

f^{'} (x) = 3 x^{2} - 16 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{16}{3},

Increasing

: \frac{16}{3} < x < \infty

Plug

x = 0

into

x^{3} - 8 x^{2} + 2 : 2

Maximum (0, 2)

Plug

x = \frac{16}{3}

into

x^{3} - 8 x^{2} + 2 : - \frac{1994}{27}

Minimum (\frac{16}{3}, - \frac{1994}{27})

Maximum (0, 2), Minimum (\frac{16}{3}, - \frac{1994}{27})

What is the extreme f(x)=x^3-8x^2+2 ?
The extreme f(x)=x^3-8x^2+2 is Maximum(0,2),Minimum(16/3 ,-1994/27)