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Precalculus : Rational Functions

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Example Questions

Example Question #1 : Rational Equations And Partial Fractions

Solve the inequality:

$1+\frac{5}{z-1}<\frac{7}{6}$

Possible Answers:

1 < z < 31

z < 1

z < 1 or z > 31

No solution

Correct answer:

z < 1 or z > 31

Explanation:

First, find the LCD, which is 6(z-1). Then, multiply the entire equation by this:

$1\cdot 6(z-1)+\frac{5}{z-1}6(z-1)<\frac{7}{6}\cdot 6(z-1)$

This simplifies to:

6z-6+30<7z-7

6z+24<7z-7

31<z

Now, make a number line with points at 1 and 31, and vertical lines in between zones:

Screen shot 2020 05 28 at 11.32.35 pm

Let's test three points, one in each zone:

f(0): $1+\frac{5}{-1} = 1-5=-6 <\frac{7}{6}$

f(10): $1+\frac{5}{10-1}=1+\frac{5}{9}=\frac{14}{9}=\frac{28}{18}\nless \frac{7}{6}$

f(35): $1+\frac{5}{35-1}=1+\frac{5}{34}=\frac{39}{34}< \frac{7}{6}$

Therefore, the inequality is true for values of z such that z < 1 or z > 31.

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Example Question #6 : Rational Equations And Partial Fractions

Decompose $\frac{3p-1}{p^2-1}$ into partial fractions.

Possible Answers:

$\frac{1}{p+1}+\frac{2}{p-1}$

$\frac{2}{p+1}+\frac{1}{p-1}$

$\frac{3p}{p+1}-\frac{1}{p-1}$

$\frac{2}{p+1}-\frac{1}{p-1}$

Correct answer:

$\frac{2}{p+1}+\frac{1}{p-1}$

Explanation:

Decomposing a fraction into partial fractions shows what fractions were added or subtracted to result in the expression we see. First, factor the denominator.

$\frac{3p-1}{p^2-1}$ $=\frac{3p-1}{(p+1)(p-1)}$

Next, we'll express the factored form as the sum of two fractions. We don't know these fractions' numerators, so we'll call them X and Y.

$\frac{3p-1}{(p+1)(p-1)}=\frac{X}{p+1}+\frac{Y}{p-1}$

Next, eliminate all of the denominators by multiplying both sides of the fraction by the LCD.

3p-1=X(p-1) + Y(p+1)

Looking back to our original fraction, note that the values p = 1 and b = -1 cannot be plugged in, as they will make the function undefined. However, if we take each of these values and plug them in, we can solve for X and Y.

First, let p=1.

3(1)-1=X(1-1) + Y(1+1)

2=2Y

1=Y

Second, let p=-1

3(-1)-1=X(-1-1) + Y(-1+1)

-4=-2X

2=X

Now, substitute these values in for X and Y.

$\frac{X}{p+1}+\frac{Y}{p-1}=\frac{2}{p+1}+\frac{1}{p-1}$

Therefore, $\frac{3p-1}{p^2-1}=\frac{2}{p+1}+\frac{1}{p-1}$ .

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Example Question #7 : Rational Equations And Partial Fractions

Decompose $\frac{2b+1}{b^2+b}$ into partial fractions.

Possible Answers:

$\frac{1}{b}-\frac{1}{b+1}$

$\frac{2b}{b}+\frac{1}{b+1}$

$\frac{1}{b}+\frac{2b}{b+1}$

$\frac{1}{b}+\frac{1}{b+1}$

Correct answer:

$\frac{1}{b}+\frac{1}{b+1}$

Explanation:

Decomposing a fraction into partial fractions shows what fractions were added or subtracted to result in the expression we see. First, factor the denominator.

$\frac{2b+1}{b^2+b}=\frac{2b+1}{b(b+1)}$

Next, we'll express the factored form as the sum of two fractions. We don't know these fractions' numerators, so we'll call them X and Y.

$\frac{2b+1}{b(b+1)}=\frac{X}{b}+\frac{Y}{b+1}$

Next, eliminate all of the denominators by multiplying both sides of the fraction by the LCD.

2b+1=X(b+1)+Yb

Looking back to our original fraction, note that the values b = 0 and b = -1 cannot be plugged in, as they will make the function undefined. However, if we take each of these values and plug them in, we can solve for X and Y.

First, let b=0.

0+1=X(0+1)+0

1=X

Second, let b=-1

2(-1)+1=X(-1+1)+Y(-1)

-2+1=-Y

-1=-Y

1=Y

Now, substitute these values in for X and Y.

$\frac{X}{b}+\frac{Y}{b+1}=\frac{1}{b}+\frac{1}{b+1}$

Therefore, $\frac{2b+1}{b^2+b}=\frac{1}{b}+\frac{1}{b+1}$ .

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Precalculus : Rational Functions

Study concepts, example questions & explanations for Precalculus

All Precalculus Resources

Example Questions

Example Question #1 : Rational Equations And Partial Fractions

Example Question #6 : Rational Equations And Partial Fractions

Example Question #7 : Rational Equations And Partial Fractions

All Precalculus Resources

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