What is the extreme 8x^2+3ln(x+1) ?

The extreme 8x^2+3ln(x+1) is Maximum(-3/4 , 9/2-6ln(2)),Minimum(-1/4 , 1/2+3ln(3/4))

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extreme 8x^2+3ln(x+1)

extreme points

8 x^{2} + 3 ln (x + 1)

Maximum (- \frac{3}{4}, \frac{9}{2} - 6 ln (2)), Minimum (- \frac{1}{4}, \frac{1}{2} + 3 ln (\frac{3}{4}))

Solution steps

f^{'} (x) = 16 x + \frac{3}{x + 1}

Find intervals:

Increasing

: - 1 < x < - \frac{3}{4},

Decreasing

: - \frac{3}{4} < x < - \frac{1}{4},

Increasing

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

Plug

x = - \frac{3}{4}

into

8 x^{2} + 3 ln (x + 1) : \frac{9}{2} - 6 ln (2)

Maximum (- \frac{3}{4}, \frac{9}{2} - 6 ln (2))

Plug

x = - \frac{1}{4}

into

8 x^{2} + 3 ln (x + 1) : \frac{1}{2} + 3 ln (\frac{3}{4})

Minimum (- \frac{1}{4}, \frac{1}{2} + 3 ln (\frac{3}{4}))

Maximum (- \frac{3}{4}, \frac{9}{2} - 6 ln (2)), Minimum (- \frac{1}{4}, \frac{1}{2} + 3 ln (\frac{3}{4}))

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

f (x) = x^{3} - 5 x^{2} + 3 x + 13

f (x, y) = x + 4 y + \frac{2}{x y}

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

f (x) = x^{3} - 3 x + 10

What is the extreme 8x^2+3ln(x+1) ?
The extreme 8x^2+3ln(x+1) is Maximum(-3/4 , 9/2-6ln(2)),Minimum(-1/4 , 1/2+3ln(3/4))