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Popular >

extreme ln(3x^2+12)

Solution

extreme points

ln (3 x^{2} + 12)

Solution

Minimum (0, ln (12))

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 (3 x^{2} + 12) : ln (12)

Minimum (0, ln (12))

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Frequently Asked Questions (FAQ)

What is the extreme ln(3x^2+12) ?
The extreme ln(3x^2+12) is Minimum(0,ln(12))