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Popular >

extreme ln(x^2+8)

Solution

extreme points

ln (x^{2} + 8)

Solution

Minimum (0, 3 ln (2))

Solution steps

f^{'} (x) = \frac{2 x}{x ^{2} + 8}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

ln (x^{2} + 8) : 3 ln (2)

Minimum (0, 3 ln (2))

Graph

Popular Examples

extreme f(x,y)=x^3+8y^3-6xy+5 extreme points

f (x, y) = x^{3} + 8 y^{3} - 6 x y + 5

extreme f(x,y)=(-3x^2-4y^2-9x+5y+3)extreme points

f (x, y) = (- 3 x^{2} - 4 y^{2} - 9 x + 5 y + 3)

extreme y=x^2e^x extreme points

y = x^{2} e^{x}

extreme f(y)=6x^2+7y^2 extreme points

f (y) = 6 x^{2} + 7 y^{2}

extreme f(x)=(t^2)/(1+t^3)extreme points

f (x) = \frac{t ^{2}}{1 + t ^{3}}

Frequently Asked Questions (FAQ)

What is the extreme ln(x^2+8) ?
The extreme ln(x^2+8) is Minimum(0,3ln(2))