What is the extreme f(x)=sqrt(x^2+10) ?

The extreme f(x)=sqrt(x^2+10) is Minimum(0,sqrt(10))

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extreme f(x)=sqrt(x^2+10)

extreme points

f (x) = x^{2} + 10

Minimum (0, 10)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

x^{2} + 10 : 10

Minimum (0, 10)

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 6 x + 7 y

f (x) = 3 x^{4} - 16 x^{3} + 18 x^{2}, - 1 \leq x \leq 4

x^{\frac{2}{3}} (6 - x)^{\frac{1}{3}}

f (x) = x^{8} e^{5 x}

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

What is the extreme f(x)=sqrt(x^2+10) ?
The extreme f(x)=sqrt(x^2+10) is Minimum(0,sqrt(10))