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

The extreme f(x)=x^3-8x^2+3 is Maximum(0,3),Minimum(16/3 ,-1967/27)

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

extreme points

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

Maximum (0, 3), Minimum (\frac{16}{3}, - \frac{1967}{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} + 3 : 3

Maximum (0, 3)

Plug

x = \frac{16}{3}

into

x^{3} - 8 x^{2} + 3 : - \frac{1967}{27}

Minimum (\frac{16}{3}, - \frac{1967}{27})

Maximum (0, 3), Minimum (\frac{16}{3}, - \frac{1967}{27})

f (x) = x^{3} - 12 x + 18

f (x) = 2 x^{3} - 20 x + 50 x - 6

f (x) = e^{- x^{2} - y^{2}}

f (x) = x - (- 2) x^{- 1}

f (x, y) = 6 x^{2} y - 9 x y

What is the extreme f(x)=x^3-8x^2+3 ?
The extreme f(x)=x^3-8x^2+3 is Maximum(0,3),Minimum(16/3 ,-1967/27)