What is the extreme f(x)= 1/2 x^2+2x-10 ?

The extreme f(x)= 1/2 x^2+2x-10 is Minimum(-2,-12)

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extreme f(x)= 1/2 x^2+2x-10

extreme points

f (x) = \frac{1}{2} x^{2} + 2 x - 10

Minimum (- 2, - 12)

Solution steps

f^{'} (x) = x + 2

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < \infty

Plug

x = - 2

into

\frac{1}{2} x^{2} + 2 x - 10 : - 12

Minimum (- 2, - 12)

\frac{1}{s ^{2} - 1}

x^{2} + 5 x - 36

f (x) = x ln (7 x), (0, \infty)

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

f (x, y) = x^{2} + 4 y^{2} - 2 x + 8 y + 4

What is the extreme f(x)= 1/2 x^2+2x-10 ?
The extreme f(x)= 1/2 x^2+2x-10 is Minimum(-2,-12)