What is the extreme f(x)=4x^3+9x^2-12x+7 ?

The extreme f(x)=4x^3+9x^2-12x+7 is Maximum(-2,35),Minimum(1/2 , 15/4)

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extreme f(x)=4x^3+9x^2-12x+7

extreme points

f (x) = 4 x^{3} + 9 x^{2} - 12 x + 7

Maximum (- 2, 35), Minimum (\frac{1}{2}, \frac{15}{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 + 7 : 35

Maximum (- 2, 35)

Plug

x = \frac{1}{2}

into

4 x^{3} + 9 x^{2} - 12 x + 7 : \frac{15}{4}

Minimum (\frac{1}{2}, \frac{15}{4})

Maximum (- 2, 35), Minimum (\frac{1}{2}, \frac{15}{4})

h (v, t) = v t - 16 t^{2}

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

f (x) = x (x + a)

x^{2} - 2 x y + 2 y

f (x) = \frac{1}{2} x^{3} - 3 x y + 3 y^{2} - x

What is the extreme f(x)=4x^3+9x^2-12x+7 ?
The extreme f(x)=4x^3+9x^2-12x+7 is Maximum(-2,35),Minimum(1/2 , 15/4)