What is the extreme y=ln(x^2)+8x+24 ?

The extreme y=ln(x^2)+8x+24 is Maximum(-1/4 ,-4ln(2)+22)

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extreme y=ln(x^2)+8x+24

extreme

Maximum (- \frac{1}{4}, - 4 ln (2) + 22)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - \frac{1}{4},

Decreasing

: - \frac{1}{4} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{1}{4}

into

ln (x^{2}) + 8 x + 24 : - 4 ln (2) + 22

Maximum (- \frac{1}{4}, - 4 ln (2) + 22)

What is the extreme y=ln(x^2)+8x+24 ?
The extreme y=ln(x^2)+8x+24 is Maximum(-1/4 ,-4ln(2)+22)