What is the extreme f(x)=3x^3-12x^2+1 ?

The extreme f(x)=3x^3-12x^2+1 is Maximum(0,1),Minimum(8/3 ,-247/9)

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extreme f(x)=3x^3-12x^2+1

extreme points

f (x) = 3 x^{3} - 12 x^{2} + 1

Maximum (0, 1), Minimum (\frac{8}{3}, - \frac{247}{9})

Solution steps

f^{'} (x) = 9 x^{2} - 24 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

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

Increasing

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

Plug

x = 0

into

3 x^{3} - 12 x^{2} + 1 : 1

Maximum (0, 1)

Plug

x = \frac{8}{3}

into

3 x^{3} - 12 x^{2} + 1 : - \frac{247}{9}

Minimum (\frac{8}{3}, - \frac{247}{9})

Maximum (0, 1), Minimum (\frac{8}{3}, - \frac{247}{9})

f (x) = (x^{2} - 1) \cdot e^{x}

f (x) = 125 x^{3} + 15 x + 7

\frac{x}{x ^{2} - x + 25}, 0 \leq x \leq 27

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

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

What is the extreme f(x)=3x^3-12x^2+1 ?
The extreme f(x)=3x^3-12x^2+1 is Maximum(0,1),Minimum(8/3 ,-247/9)