What is the extreme f(x)=(8x^3)/3-5x^2-3x+4 ?

The extreme f(x)=(8x^3)/3-5x^2-3x+4 is Maximum(-1/4 , 211/48),Minimum(3/2 ,-11/4)

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extreme f(x)=(8x^3)/3-5x^2-3x+4

extreme points

f (x) = \frac{8 x ^{3}}{3} - 5 x^{2} - 3 x + 4

Maximum (- \frac{1}{4}, \frac{211}{48}), Minimum (\frac{3}{2}, - \frac{11}{4})

Solution steps

f^{'} (x) = 8 x^{2} - 10 x - 3

Find intervals:

Increasing

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

Decreasing

: - \frac{1}{4} < x < \frac{3}{2},

Increasing

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

Plug

x = - \frac{1}{4}

into

\frac{8 x ^{3}}{3} - 5 x^{2} - 3 x + 4 : \frac{211}{48}

Maximum (- \frac{1}{4}, \frac{211}{48})

Plug

x = \frac{3}{2}

into

\frac{8 x ^{3}}{3} - 5 x^{2} - 3 x + 4 : - \frac{11}{4}

Minimum (\frac{3}{2}, - \frac{11}{4})

Maximum (- \frac{1}{4}, \frac{211}{48}), Minimum (\frac{3}{2}, - \frac{11}{4})

f (x, y) = 6 x^{2} - 2 x^{3} + 3 y^{2} + 6 x y

g (x, y) = x^{3} - 3 x y^{2} + y^{3}

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

f (x) = \frac{x}{x ^{2} - x + 4}, 0 \leq x \leq 6

f (x) = x^{3} - 6 x^{2} + 11

What is the extreme f(x)=(8x^3)/3-5x^2-3x+4 ?
The extreme f(x)=(8x^3)/3-5x^2-3x+4 is Maximum(-1/4 , 211/48),Minimum(3/2 ,-11/4)