What is the extreme f(x)=-10/3 x^3-31/2 x^2-3x+4 ?

The extreme f(x)=-10/3 x^3-31/2 x^2-3x+4 is Minimum(-3,-73/2),Maximum(-1/10 , 2489/600)

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extreme f(x)=-10/3 x^3-31/2 x^2-3x+4

extreme points

f (x) = - \frac{10}{3} x^{3} - \frac{31}{2} x^{2} - 3 x + 4

Minimum (- 3, - \frac{73}{2}), Maximum (- \frac{1}{10}, \frac{2489}{600})

Solution steps

f^{'} (x) = - 10 x^{2} - 31 x - 3

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < - \frac{1}{10},

Decreasing

: - \frac{1}{10} < x < \infty

Plug

x = - 3

into

- \frac{10}{3} x^{3} - \frac{31}{2} x^{2} - 3 x + 4 : - \frac{73}{2}

Minimum (- 3, - \frac{73}{2})

Plug

x = - \frac{1}{10}

into

- \frac{10}{3} x^{3} - \frac{31}{2} x^{2} - 3 x + 4 : \frac{2489}{600}

Maximum (- \frac{1}{10}, \frac{2489}{600})

Minimum (- 3, - \frac{73}{2}), Maximum (- \frac{1}{10}, \frac{2489}{600})

f (x) = \frac{x ^{2} + 7 x + 13}{x + 4}

\frac{( 5 x - 8 )}{x}

f (x) = sin (x), - 5 \leq x \leq 6

f (x) = \frac{- 2 x ^{2} + 16}{16 - y ^{2}}

f (x) = 8 + (9 + 5 x)^{\frac{2}{5}}

What is the extreme f(x)=-10/3 x^3-31/2 x^2-3x+4 ?
The extreme f(x)=-10/3 x^3-31/2 x^2-3x+4 is Minimum(-3,-73/2),Maximum(-1/10 , 2489/600)