What is the extreme f(x)=x^3+3/2 x^2-6x+10 ?

The extreme f(x)=x^3+3/2 x^2-6x+10 is Maximum(-2,20),Minimum(1, 13/2)

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extreme f(x)=x^3+3/2 x^2-6x+10

extreme points

f (x) = x^{3} + \frac{3}{2} x^{2} - 6 x + 10

Maximum (- 2, 20), Minimum (1, \frac{13}{2})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 2

into

x^{3} + \frac{3}{2} x^{2} - 6 x + 10 : 20

Maximum (- 2, 20)

Plug

x = 1

into

x^{3} + \frac{3}{2} x^{2} - 6 x + 10 : \frac{13}{2}

Minimum (1, \frac{13}{2})

Maximum (- 2, 20), Minimum (1, \frac{13}{2})

f (x) = 4^{x - 2}

\frac{x}{( x + 2 ) ( x - 3 )}

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

(0) (6 - 2)

f (x) = 7 - 16 t

What is the extreme f(x)=x^3+3/2 x^2-6x+10 ?
The extreme f(x)=x^3+3/2 x^2-6x+10 is Maximum(-2,20),Minimum(1, 13/2)