decreasing function

**Samantha** · August 1st 2007, 04:23 AM

Find all open intervals on which the function $f(x) = \frac {x} {x^2 + x - 2}$ is decreasing.

**Samantha** · August 1st 2007, 04:27 AM

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

$<br /> 1(x^2+x-2)-(2x+1)(x)<br />$
$<br /> x^2+x-2-2x^2-x=-x^2-2$

f(prime)(x)= $\frac {-(x^2+2)}[\math] {((x+2)(x-1))^2<br />$

**topsquark** · August 1st 2007, 04:29 AM

Originally Posted by Samantha

Find all open intervals on which the function f(x) = \frac {x} {x^2 + x - 2} is decreasing.

$f^{\prime}(x) = \frac{1 \cdot (x^2 + x - 2) - x \cdot (2x + 1)}{(x^2 + x - 2)^2}$

$f^{\prime}(x) = \frac{-x^2 - 2}{(x^2 + x - 2)^2}$

When is this negative? (Hint: The denominator is always positive, since it is squared.)

-Dan

**Samantha** · August 1st 2007, 04:37 AM

So it's always decreasing?

**topsquark** · August 1st 2007, 05:01 AM

Originally Posted by Samantha

So it's always decreasing?

It looks like it.

-Dan

**Samantha** · August 1st 2007, 05:06 AM

so it's (-infinity, -2), (-2,1) and (1, infinity)?

**topsquark** · August 1st 2007, 12:25 PM

Originally Posted by Samantha

so it's (-infinity, -2), (-2,1) and (1, infinity)?

Yup. (You are more exact than I was going to be. I probably wouldn't have removed the points that are outside the domain, and you were right to do so. I would've lost points on this problem.

)

-Dan

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