What is the extreme f(x)=8+6x-7x^2-4x^3 ?

The extreme f(x)=8+6x-7x^2-4x^3 is Minimum(-3/2 ,-13/4),Maximum(1/3 , 245/27)

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extreme f(x)=8+6x-7x^2-4x^3

extreme points

f (x) = 8 + 6 x - 7 x^{2} - 4 x^{3}

Minimum (- \frac{3}{2}, - \frac{13}{4}), Maximum (\frac{1}{3}, \frac{245}{27})

Solution steps

f^{'} (x) = - 12 x^{2} - 14 x + 6

Find intervals:

Decreasing

: - \infty < x < - \frac{3}{2},

Increasing

: - \frac{3}{2} < x < \frac{1}{3},

Decreasing

: \frac{1}{3} < x < \infty

Plug

x = - \frac{3}{2}

into

8 + 6 x - 7 x^{2} - 4 x^{3} : - \frac{13}{4}

Minimum (- \frac{3}{2}, - \frac{13}{4})

Plug

x = \frac{1}{3}

into

8 + 6 x - 7 x^{2} - 4 x^{3} : \frac{245}{27}

Maximum (\frac{1}{3}, \frac{245}{27})

Minimum (- \frac{3}{2}, - \frac{13}{4}), Maximum (\frac{1}{3}, \frac{245}{27})

f (x) = \frac{20}{3 ( 2 + x ) ^{3}}

f (x) = \frac{- x - 5}{x ^{2} + 16 x + 64}

y = 2 sin (∣ x ∣), - 2 π \leq x \leq 2 π

f (x) = x^{3} \cdot ln (x)

f (x) = x^{3} + 3 x^{2} + 5, - 3 \leq x \leq 2

What is the extreme f(x)=8+6x-7x^2-4x^3 ?
The extreme f(x)=8+6x-7x^2-4x^3 is Minimum(-3/2 ,-13/4),Maximum(1/3 , 245/27)