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Popular >

extreme f(x)=sqrt(x^2+16)

Solution

extreme points

f (x) = x^{2} + 16

Solution

Minimum (0, 4)

Solution steps

f^{'} (x) = \frac{x}{x ^{2} + 16}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

x^{2} + 16 : 4

Minimum (0, 4)

Graph

Popular Examples

extreme f(x)=x^2-6x-9 extreme points

f (x) = x^{2} - 6 x - 9

extreme f(x)=2x^3+18x^2-42x+7,-7<= x<= 2 extreme points

f (x) = 2 x^{3} + 18 x^{2} - 42 x + 7, - 7 \leq x \leq 2

extreme f(x)=(x^2+2)/(x^2-25)extreme points

f (x) = \frac{x ^{2} + 2}{x ^{2} - 25}

extreme f(x)=(4e^{4x})/(4x-5)extreme points

f (x) = \frac{4 e ^{4 x}}{4 x - 5}

extreme f(x)=sqrt(x+3)+6 extreme points

f (x) = x + 3 + 6

Frequently Asked Questions (FAQ)

What is the extreme f(x)=sqrt(x^2+16) ?
The extreme f(x)=sqrt(x^2+16) is Minimum(0,4)