What is the extreme f(x)=(x^3)/2-2x^2+1 ?

The extreme f(x)=(x^3)/2-2x^2+1 is Maximum(0,1),Minimum(8/3 ,-101/27)

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extreme f(x)=(x^3)/2-2x^2+1

extreme points

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

Maximum (0, 1), Minimum (\frac{8}{3}, - \frac{101}{27})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{8}{3},

Increasing

: \frac{8}{3} < x < \infty

Plug

x = 0

into

\frac{x ^{3}}{2} - 2 x^{2} + 1 : 1

Maximum (0, 1)

Plug

x = \frac{8}{3}

into

\frac{x ^{3}}{2} - 2 x^{2} + 1 : - \frac{101}{27}

Minimum (\frac{8}{3}, - \frac{101}{27})

Maximum (0, 1), Minimum (\frac{8}{3}, - \frac{101}{27})

3 t (8 - t)

f (x) = - x^{2} + 50 ln (x)

f (x) = - 69.7 x^{2} + 625.6 x

f (x) = \frac{1}{2} (x + 4)^{2} + 6

f (x) = - 0.3 x^{2} + 2.4 x + 98.8

What is the extreme f(x)=(x^3)/2-2x^2+1 ?
The extreme f(x)=(x^3)/2-2x^2+1 is Maximum(0,1),Minimum(8/3 ,-101/27)