What is the extreme f(x)=7+x-x^2-ln((x+3)^3) ?

The extreme f(x)=7+x-x^2-ln((x+3)^3) is Minimum(-5/2 ,3ln(2)-7/4),Maximum(0,-3ln(3)+7)

Popular >

extreme f(x)=7+x-x^2-ln((x+3)^3)

extreme points

f (x) = 7 + x - x^{2} - ln ((x + 3)^{3})

Minimum (- \frac{5}{2}, 3 ln (2) - \frac{7}{4}), Maximum (0, - 3 ln (3) + 7)

Solution steps

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

Find intervals:

Decreasing

: - 3 < x < - \frac{5}{2},

Increasing

: - \frac{5}{2} < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - \frac{5}{2}

into

7 + x - x^{2} - ln ((x + 3)^{3}) : 3 ln (2) - \frac{7}{4}

Minimum (- \frac{5}{2}, 3 ln (2) - \frac{7}{4})

Plug

x = 0

into

7 + x - x^{2} - ln ((x + 3)^{3}) : - 3 ln (3) + 7

Maximum (0, - 3 ln (3) + 7)

Minimum (- \frac{5}{2}, 3 ln (2) - \frac{7}{4}), Maximum (0, - 3 ln (3) + 7)

(x^{2} - 1)^{3}, - 1 \leq x \leq 3

f (x) = 6 x^{2} - 7 x - 11

f (x, y) = (x^{2} + x y + y^{2} - 19 y + 120)

f (x, y) = \frac{1}{2} (x^{3} - x - y^{2} + 3)

f (x) = \frac{( 5 x )}{( x ^{2} + 5 )}

What is the extreme f(x)=7+x-x^2-ln((x+3)^3) ?
The extreme f(x)=7+x-x^2-ln((x+3)^3) is Minimum(-5/2 ,3ln(2)-7/4),Maximum(0,-3ln(3)+7)