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Popular >

extreme (5x^2)/(x^2+16)

Solution

extreme points

\frac{5 x ^{2}}{x ^{2} + 16}

Solution

Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{160 x}{( x ^{2} + 16 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

\frac{5 x ^{2}}{x ^{2} + 16} : 0

Minimum (0, 0)

Popular Examples

extreme f(x)=8x-x^4,-2<= x<= 1 extreme points

f (x) = 8 x - x^{4}, - 2 \leq x \leq 1

extreme f(x,y)=2x^2+4xy^2-2xy-10x-2y+2 extreme points

f (x, y) = 2 x^{2} + 4 x y^{2} - 2 x y - 10 x - 2 y + 2

extreme f(x)=3(x-4)(x+7)^2 extreme points

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

extreme f(x)=x^3+2x^2+x extreme points

f (x) = x^{3} + 2 x^{2} + x

extreme x^3-3x^2+3x+1 extreme points

x^{3} - 3 x^{2} + 3 x + 1

Frequently Asked Questions (FAQ)

What is the extreme (5x^2)/(x^2+16) ?
The extreme (5x^2)/(x^2+16) is Minimum(0,0)