Re: Rational Inequalities

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**tdotodot** 3. Now I look for the "roots" or "zeros" for numerator and denominator?

....N: x can't be ____?

....D: x can't be __+3__.

Solve the numerator for 0 anyway. The values this gives are also "critical points" for your intervals. (They typically are points where the function goes from + to - or - to +.)

-Dan

Re: Rational Inequalities

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**topsquark** Solve the numerator for 0 anyway. The values this gives are also "critical points" for your intervals. (They typically are points where the function goes from + to - or - to +.)

-Dan

Okay, here's what I did.

Numerator = 0 ->x=4.8 and x=0.2

Denominator = 0 -> x=3

Number line: |-----0.2-------3-------4.8-------|

Checked each number: 0.2 works , 3 does not work , 4.8 works.

Tested values in between each interval.

f(0.1) <=0

f(1) <=0

f(4) <=0

f(5) <=0

All integers tested between each interval were less than or equal to zero.

So then my final answer would be: (-infiniti,2) U (4, +infiniti)

Is that right?

Re: Rational Inequalities

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**tdotodot** Solve each inequality by graphing, using technology. Express your answer to one decimal place.

a) 2x+1/x-3 <= x

Okay, I need help solving this. Am I on the right track?

1. Make equal to zero : x- 2x+1/x-3 <= 0

This is not quite right. You had $\displaystyle \dfrac{2x+1}{x-3} \le x$. Subtract $\displaystyle \dfrac{2x+1}{x-3}$ from each side to get $\displaystyle 0 \le x - \dfrac{2x+1}{x-3}$. Your inequality is flipped the other way.

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**tdotodot** 2. "combine terms"? : x(x-3)-2x+1 / x-3 <= 0

............................. x^2-5x+1 / x-3 ........up to here is this correct?

Not quite. $\displaystyle x-\dfrac{2x+1}{x-3} = \dfrac{x(x-3) - (2x+1)}{x-3} = \dfrac{x^2-3x-2x-1}{x-3} = \dfrac{x^2-5x-1}{x-3}$. You had $\displaystyle x^2-5x+1$ in the numerator.

Re: Rational Inequalities

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**SlipEternal** This is not quite right. You had $\displaystyle \dfrac{2x+1}{x-3} \le x$. Subtract $\displaystyle \dfrac{2x+1}{x-3}$ from each side to get $\displaystyle 0 \le x - \dfrac{2x+1}{x-3}$. Your inequality is flipped the other way.

Not quite. $\displaystyle x-\dfrac{2x+1}{x-3} = \dfrac{x(x-3) - (2x+1)}{x-3} = \dfrac{x^2-3x-2x-1}{x-3} = \dfrac{x^2-5x-1}{x-3}$. You had $\displaystyle x^2-5x+1$ in the numerator.

Okay, so here is what I get now...

Numerator = 0 -> x=5.2 and x=-0.2

Denominator = 0 ->x=3

|--------(-0.2)-----------(3)----------(5.2)-----|

Test Values: (-0.2) = -0.0125 ... is Not greater than or equal to zero. *I am plugging this into "0 <= x^2-5x-1 / x-3"*

................... (3) = undefined

....................(5.2) = greater than or equal to zero.

Then pick values between each interval:

f(-1) -> 1.25 which is >= 0

f(1) -> 2.5 which is >= 0

f(4) -> -5 which is < 0

f(6) ->1.67 which is >= 0

So back to the original equation now?? ... 2x+1 / x-3 <= 0

.................................................. ......when is it less than or equal to zero?

.................................................. ......(4, 5.1)?

.................................................. ......I'm guessing the answer is wrong? Can someone please explain this to me? :)

Re: Rational Inequalities

Quote:

Originally Posted by

**tdotodot** Solve each inequality by graphing, using technology. Express your answer to one decimal place.

a) 2x+1/x-3 <= x.

Look at this solution.