What is the extreme f(x)=20x^3+9x^2-24x+12 ?

The extreme f(x)=20x^3+9x^2-24x+12 is Maximum(-4/5 , 668/25),Minimum(1/2 , 19/4)

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extreme f(x)=20x^3+9x^2-24x+12

extreme points

f (x) = 20 x^{3} + 9 x^{2} - 24 x + 12

Maximum (- \frac{4}{5}, \frac{668}{25}), Minimum (\frac{1}{2}, \frac{19}{4})

Solution steps

f^{'} (x) = 60 x^{2} + 18 x - 24

Find intervals:

Increasing

: - \infty < x < - \frac{4}{5},

Decreasing

: - \frac{4}{5} < x < \frac{1}{2},

Increasing

: \frac{1}{2} < x < \infty

Plug

x = - \frac{4}{5}

into

20 x^{3} + 9 x^{2} - 24 x + 12 : \frac{668}{25}

Maximum (- \frac{4}{5}, \frac{668}{25})

Plug

x = \frac{1}{2}

into

20 x^{3} + 9 x^{2} - 24 x + 12 : \frac{19}{4}

Minimum (\frac{1}{2}, \frac{19}{4})

Maximum (- \frac{4}{5}, \frac{668}{25}), Minimum (\frac{1}{2}, \frac{19}{4})

f (x) = x^{2} + 9 x + 14

f (x) = - x^{2} - 5 x + 1

f (x) = \frac{x ^{4}}{4} - \frac{4 x ^{3}}{3} + 2 x^{2} - 1

- 30 x^{2} - 25 y^{2} + 480 x + 300 y + 170

f (x) = \frac{( x - 2 ) ^{2}}{x - 5}

What is the extreme f(x)=20x^3+9x^2-24x+12 ?
The extreme f(x)=20x^3+9x^2-24x+12 is Maximum(-4/5 , 668/25),Minimum(1/2 , 19/4)