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Popular >

extreme f(x)=ln(1+x^2)

Solution

extreme points

f (x) = ln (1 + x^{2})

Solution

Minimum (0, 0)

Solution steps

f^{'} (x) = \frac{2 x}{1 + x ^{2}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

ln (1 + x^{2}) : 0

Minimum (0, 0)

Popular Examples

extreme-3x^5+5x^3 extreme points

- 3 x^{5} + 5 x^{3}

extreme f(x,y)=x^2+y^2-x-y+1 extreme points

f (x, y) = x^{2} + y^{2} - x - y + 1

extreme f(x,y)=(x-y)(16-xy)extreme points

f (x, y) = (x - y) (16 - x y)

extreme f(x,y)=(1+xy)(x+y)extreme points

f (x, y) = (1 + x y) (x + y)

extreme f(x)=3x^4-4x^3-12x^2+1 extreme points

f (x) = 3 x^{4} - 4 x^{3} - 12 x^{2} + 1

Frequently Asked Questions (FAQ)

What is the extreme f(x)=ln(1+x^2) ?
The extreme f(x)=ln(1+x^2) is Minimum(0,0)