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Popular >

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

Solution

extreme points

f (x) = ln (x^{2} + 4)

Solution

Minimum (0, 2 ln (2))

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

ln (x^{2} + 4) : 2 ln (2)

Minimum (0, 2 ln (2))

Graph

Popular Examples

extreme f(x)=x(25-40+2x)(40/2-x)extreme points

f (x) = x (25 - 40 + 2 x) (\frac{40}{2} - x)

extreme f(x)=x^{1/5}(x+12)extreme points

f (x) = x^{\frac{1}{5}} (x + 12)

extreme F(x,y)=x^2y^2+x^2y+xy^2+xy extreme points

F (x, y) = x^{2} y^{2} + x^{2} y + x y^{2} + x y

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f (x) = ((x - 6)^{(\frac{4}{7})}) - 3

extreme x^3-8y^3+4xy-8 extreme points

x^{3} - 8 y^{3} + 4 x y - 8

Frequently Asked Questions (FAQ)

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