What is the extreme f(x)= 10/3 x^3-x^2-8x+48 ?

The extreme f(x)= 10/3 x^3-x^2-8x+48 is Maximum(-4/5 , 3904/75),Minimum(1, 127/3)

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extreme f(x)= 10/3 x^3-x^2-8x+48

extreme points

f (x) = \frac{10}{3} x^{3} - x^{2} - 8 x + 48

Maximum (- \frac{4}{5}, \frac{3904}{75}), Minimum (1, \frac{127}{3})

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 1 < x < \infty

Plug

x = - \frac{4}{5}

into

\frac{10}{3} x^{3} - x^{2} - 8 x + 48 : \frac{3904}{75}

Maximum (- \frac{4}{5}, \frac{3904}{75})

Plug

x = 1

into

\frac{10}{3} x^{3} - x^{2} - 8 x + 48 : \frac{127}{3}

Minimum (1, \frac{127}{3})

Maximum (- \frac{4}{5}, \frac{3904}{75}), Minimum (1, \frac{127}{3})

f (x, y) = x y - 2 x - y

f (x) = x^{2} - 6 x - 7

f (x) = x e^{5 x^{2}}

f (x, y) = 8 x^{2} + 14 x y + 3 y^{2} + 10 x - 4

(x^{2} + 10) (9 - x^{2})

What is the extreme f(x)= 10/3 x^3-x^2-8x+48 ?
The extreme f(x)= 10/3 x^3-x^2-8x+48 is Maximum(-4/5 , 3904/75),Minimum(1, 127/3)