What is the extreme f(x)=((4x^3))/3-4x^2-60x+35 ?

The extreme f(x)=((4x^3))/3-4x^2-60x+35 is Maximum(-3,143),Minimum(5,-595/3)

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extreme f(x)=((4x^3))/3-4x^2-60x+35

extreme points

f (x) = \frac{( 4 x ^{3} )}{3} - 4 x^{2} - 60 x + 35

Maximum (- 3, 143), Minimum (5, - \frac{595}{3})

Solution steps

f^{'} (x) = 4 x^{2} - 8 x - 60

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = - 3

into

\frac{4 x ^{3}}{3} - 4 x^{2} - 60 x + 35 : 143

Maximum (- 3, 143)

Plug

x = 5

into

\frac{4 x ^{3}}{3} - 4 x^{2} - 60 x + 35 : - \frac{595}{3}

Minimum (5, - \frac{595}{3})

Maximum (- 3, 143), Minimum (5, - \frac{595}{3})

f (x, y) = 12 x y + 4321

f (x) = 5 x + 9

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

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

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

What is the extreme f(x)=((4x^3))/3-4x^2-60x+35 ?
The extreme f(x)=((4x^3))/3-4x^2-60x+35 is Maximum(-3,143),Minimum(5,-595/3)