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Popular >

extreme y=ln(x^2+2)

Solution

extreme points

y = ln (x^{2} + 2)

Solution

Minimum (0, ln (2))

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

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

Minimum (0, ln (2))

Popular Examples

extreme f(x)=e^{2x}(x^2+2x)extreme points

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

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

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f (x) = \frac{7 x}{x ^{2} + 4}

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

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

Frequently Asked Questions (FAQ)

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