Find all open intervals on which the function $\displaystyle f(x) = \frac {x} {x^2 + x - 2}$ is decreasing.
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f(x) = $\displaystyle \frac {x} {x^2+x-2}= \frac {x} {(x+2)(x-1)}$ $\displaystyle 1(x^2+x-2)-(2x+1)(x) $ $\displaystyle x^2+x-2-2x^2-x=-x^2-2 $ f(prime)(x)= $\displaystyle \frac {-(x^2+2)}[\math] {((x+2)(x-1))^2 $
Originally Posted by Samantha Find all open intervals on which the function f(x) = \frac {x} {x^2 + x - 2} is decreasing. $\displaystyle f^{\prime}(x) = \frac{1 \cdot (x^2 + x - 2) - x \cdot (2x + 1)}{(x^2 + x - 2)^2}$ $\displaystyle f^{\prime}(x) = \frac{-x^2 - 2}{(x^2 + x - 2)^2}$ When is this negative? (Hint: The denominator is always positive, since it is squared.) -Dan
So it's always decreasing?
Originally Posted by Samantha So it's always decreasing? It looks like it. -Dan
so it's (-infinity, -2), (-2,1) and (1, infinity)?
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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