What is the extreme f(x)=4x^3+9x^2-12x+5 ?

The extreme f(x)=4x^3+9x^2-12x+5 is Maximum(-2,33),Minimum(1/2 , 7/4)

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extreme f(x)=4x^3+9x^2-12x+5

extreme points

f (x) = 4 x^{3} + 9 x^{2} - 12 x + 5

Maximum (- 2, 33), Minimum (\frac{1}{2}, \frac{7}{4})

Solution steps

f^{'} (x) = 12 x^{2} + 18 x - 12

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

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

Increasing

: \frac{1}{2} < x < \infty

Plug

x = - 2

into

4 x^{3} + 9 x^{2} - 12 x + 5 : 33

Maximum (- 2, 33)

Plug

x = \frac{1}{2}

into

4 x^{3} + 9 x^{2} - 12 x + 5 : \frac{7}{4}

Minimum (\frac{1}{2}, \frac{7}{4})

Maximum (- 2, 33), Minimum (\frac{1}{2}, \frac{7}{4})

f (x) = x^{3} - 27 x + 5

f (x) = 6 x + \frac{384}{x}

f (x) = - 2 x^{3} + 36 x^{2} - 192 x

f (x) = 2 x^{3} - 6 x^{2} - 144 x + 7

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

What is the extreme f(x)=4x^3+9x^2-12x+5 ?
The extreme f(x)=4x^3+9x^2-12x+5 is Maximum(-2,33),Minimum(1/2 , 7/4)