What is the extreme 40x^3+9x^2-12x-8 ?

The extreme 40x^3+9x^2-12x-8 is Maximum(-2/5 ,-108/25),Minimum(1/4 ,-157/16)

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extreme 40x^3+9x^2-12x-8

extreme points

40 x^{3} + 9 x^{2} - 12 x - 8

Maximum (- \frac{2}{5}, - \frac{108}{25}), Minimum (\frac{1}{4}, - \frac{157}{16})

Solution steps

f^{'} (x) = 120 x^{2} + 18 x - 12

Find intervals:

Increasing

: - \infty < x < - \frac{2}{5},

Decreasing

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

Increasing

: \frac{1}{4} < x < \infty

Plug

x = - \frac{2}{5}

into

40 x^{3} + 9 x^{2} - 12 x - 8 : - \frac{108}{25}

Maximum (- \frac{2}{5}, - \frac{108}{25})

Plug

x = \frac{1}{4}

into

40 x^{3} + 9 x^{2} - 12 x - 8 : - \frac{157}{16}

Minimum (\frac{1}{4}, - \frac{157}{16})

Maximum (- \frac{2}{5}, - \frac{108}{25}), Minimum (\frac{1}{4}, - \frac{157}{16})

f (x, y) = (x^{2} + 3 y^{2}) e^{1 - x^{2} - y^{2}}

f (y) = - x^{3} - 5 x^{2} - 7 x + 3

f (x) = \frac{7 - x}{6 + x}

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

x^{3} - 8 x^{2} + 16 x

What is the extreme 40x^3+9x^2-12x-8 ?
The extreme 40x^3+9x^2-12x-8 is Maximum(-2/5 ,-108/25),Minimum(1/4 ,-157/16)