What is the extreme f(x)=(4x^2)/2+(x^2)/2 ?

The extreme f(x)=(4x^2)/2+(x^2)/2 is Minimum(0,0)

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extreme f(x)=(4x^2)/2+(x^2)/2

extreme points

f (x) = \frac{4 x ^{2}}{2} + \frac{x ^{2}}{2}

Minimum (0, 0)

Solution steps

f^{'} (x) = 5 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

\frac{4 x ^{2}}{2} + \frac{x ^{2}}{2} : 0

Minimum (0, 0)

f (r, h) = 2 π r h + 2 π r^{2}

g (t) = u (t)

f (x) = x^{3} + 10 x - 7

f (x) = 9 x - 18 cos (x)

What is the extreme f(x)=(4x^2)/2+(x^2)/2 ?
The extreme f(x)=(4x^2)/2+(x^2)/2 is Minimum(0,0)