Find zero/Find positive intervals

**cam_ddrt** · May 29th 2009, 05:56 PM

I am studying for a test and I cannot figure out how to do these two questions. Any help would be greatly appreciated. Thanks!

1. Find the intervals on which (x²-4)/(x-1) is positive

2. Find all of x for which (x²-x)(2x+1) - (x²+x)(2x-1) is zero

**Soroban** · May 29th 2009, 07:22 PM

Hello, cam_ddrt!

1. Find the intervals on which $f(x) \:=\:\frac{x^2-4}{x-1}$ is positive.

We have: . $f(x) \:=\:\frac{(x-2)(x+2)}{x-1}$

The function is zero or undefined at $x \:=\:-2, 1, 2$
These are the only values where $f(x)$ changes signs.

These three values divide the number line into four intervals.

. . $\underbrace{\;- - - -}\;[\text{-}2]\;\underbrace{ - - - }\;[1]\;\underbrace{ - - - }\;[2]\;\underbrace{ - - -\;}$

Test a value in each interval.

. . $f(\text{-}3) \:=\:\frac{(\text{-}5)(\text{-}1)}{\text{-}3-1}\:=\:\frac{5}{\text{-}4}\;\text{ neg.}$

. . $f(0) \:=\:\frac{0-4}{0-1} \:=\:4\;\text{ pos.}$

. . $f(\tfrac{3}{2}) \:=\:\frac{(-\frac{1}{2})(\frac{7}{2})}{\frac{1}{2}} \:=\: \text{-}\frac{7}{2}\;\text{ neg.}$

. . $f(3) \:=\:\frac{(1)(5)}{(2} \:=\:\frac{5}{2}\;\text{ pos.}$

We have:

. . $\underbrace{\;- - - -}_{{\color{red}-}}\;[\text{-}2]\;\underbrace{ - - - }_{{\color{red}+}}\;[1]\;\underbrace{ - - - }_{{\color{red}-}}\;[2]\;\underbrace{ - - -\;}_{{\color{red}+}}$

Therefore, $f(x)$ is positive on: . $(\text{-}2,1) \cup (2,\infty)$

2. Find all $x$ for which $f(x) \:=\:(x^2-x)(2x+1)-(x^2+x)(2x-1)$ is zero.

We have: . $f(x) \:=\:0$

Simplify: . $(x^2-x)(2x+1) - (x^2+x)(2x-1) \:=\:0$

. . . . . . . . . $(2x^3 - x^2 - x) - (2x^3 + x^2 - x) \:=\:0$

. . . . . . . . . . . $2x^3 - x^2 - x - 2x^3 - x^2 + x \:=\:0$

. . . . . . . . . . . . . . . . . . . . . . . . . $-2x^2 \:=\:0$

. . . . . . . . . . . . . . . . . . . . . . . . . . . $x \:=\:0$

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