What is the extreme f(x)=4x^3-3x^2-60x+17 ?

The extreme f(x)=4x^3-3x^2-60x+17 is Maximum(-2,93),Minimum(5/2 ,-357/4)

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extreme f(x)=4x^3-3x^2-60x+17

extreme points

f (x) = 4 x^{3} - 3 x^{2} - 60 x + 17

Maximum (- 2, 93), Minimum (\frac{5}{2}, - \frac{357}{4})

Solution steps

f^{'} (x) = 12 x^{2} - 6 x - 60

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < \frac{5}{2},

Increasing

: \frac{5}{2} < x < \infty

Plug

x = - 2

into

4 x^{3} - 3 x^{2} - 60 x + 17 : 93

Maximum (- 2, 93)

Plug

x = \frac{5}{2}

into

4 x^{3} - 3 x^{2} - 60 x + 17 : - \frac{357}{4}

Minimum (\frac{5}{2}, - \frac{357}{4})

Maximum (- 2, 93), Minimum (\frac{5}{2}, - \frac{357}{4})

\frac{1}{x ^{2} + 3}

f (x) = 2 x^{4}

f (x) = x x + 3

\frac{1}{x - 3}

\frac{2}{2 - x}

What is the extreme f(x)=4x^3-3x^2-60x+17 ?
The extreme f(x)=4x^3-3x^2-60x+17 is Maximum(-2,93),Minimum(5/2 ,-357/4)