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Popular >

extreme f(x)= 1/(x^2-9)

Solution

extreme points

f (x) = \frac{1}{x ^{2} - 9}

Solution

Maximum (0, - \frac{1}{9})

Solution steps

f^{'} (x) = - \frac{2 x}{( x ^{2} - 9 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = 0

into

\frac{1}{x ^{2} - 9} : - \frac{1}{9}

Maximum (0, - \frac{1}{9})

Popular Examples

extreme y=2x^3-3x+5 extreme points

y = 2 x^{3} - 3 x + 5

extreme (x^2-3)/(x+2)extreme points

\frac{x ^{2} - 3}{x + 2}

extreme f(x)= 1/(x^2+y^2+1)extreme points

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

extreme f(x)= 1/2 x^4-2x^3-54x^2+15 extreme points

f (x) = \frac{1}{2} x^{4} - 2 x^{3} - 54 x^{2} + 15

extreme f(x)=x^{1/3}(8-x)extreme points

f (x) = x^{\frac{1}{3}} (8 - x)

Frequently Asked Questions (FAQ)

What is the extreme f(x)= 1/(x^2-9) ?
The extreme f(x)= 1/(x^2-9) is Maximum(0,-1/9)