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Precalculus Review Materials

Jane Butterfield, Gary MacGillivray

Section 2.6 Exercises

These are the exercises in Module 2.
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Checkpoint 2.6.1.

Checkpoint 2.6.2.

Checkpoint 2.6.3.

Checkpoint 2.6.4.

The domain of the function
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\(f(x)=\sqrt{x(x-14)}\)
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Write the answer in interval notation.
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Note: If the answer includes more than one interval, write the intervals seperated by the union symbol, U. If needed end \(-\infty\) as -infinity and \(\infty\) as infinity.
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Answer.
\(\left(-\infty ,0\right]\cup \left[14,\infty \right)\)
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Checkpoint 2.6.5.

Checkpoint 2.6.6.

Checkpoint 2.6.7.

Checkpoint 2.6.8.

Checkpoint 2.6.9.

Select the expression that is equivalent to
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\(\frac{x-1}{5x}\cdot\frac{x}{x^2-1}\)
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Answer.
\(\text{Choice 2}\)
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Checkpoint 2.6.10.

Write \(\frac{m}{2}+\frac{m}{3}\) as a single fraction.
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Answer.
\(\text{Choice 4}\)
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Checkpoint 2.6.11.

Select the expression that is equivalent to
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\(\frac{x-2}{5x}\cdot\frac{x}{x^2-4}\)
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Answer.
\(\text{Choice 4}\)
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Checkpoint 2.6.12.

Checkpoint 2.6.13.

Evaluate the following expression without using a calculator. Simplify your answer as much as possible, and enter your answer as a fraction.
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\(\displaystyle \left( \frac{27}{8} \right)^{-1/3}\) =
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Answer.
\({\frac{2}{3}}\)
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Checkpoint 2.6.14.

Simplify and write the following as a product of rational powers of \(p\) and \(z\text{.}\)
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If \(\displaystyle \frac{ \left(p^5 z^4\right)^{1/15}} {p^{-1/5}z^{1/3}} = p^m z^n\) then \(m =\) and \(n =\)
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Answer 1.
\({\frac{8}{15}}\)
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Answer 2.
\(-{\frac{1}{15}}\)
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Solution.
\(\displaystyle{\begin{aligned} \amp \displaystyle \frac{ \left(p^5 z^4\right)^{1/15}} {p^{-1/5}z^{1/3}} = \frac{ p^{5/15} z^{4/15} }{p^{-1/5} z^{1/3}} \\ \amp = p^{(5/15+1/5)}z^{(4/15-1/3)}=p^{{{\frac{8}{15}}}}z^{{-{\frac{1}{15}}}}\end{aligned}}\)
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Checkpoint 2.6.15.

Checkpoint 2.6.16.

The expression
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\begin{equation*} \frac{2^52^42^{-5}}{\sqrt{4^2}2^32^{-5}} \end{equation*}
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equals \(2^n\) where \(n\) is:
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Answer.
\(4\)
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Solution.
\(\begin{aligned} \amp \phantom{\mathrel{{}={}}}\frac{2^{5}2^{4}2^{-5} }{ \sqrt{4^{2}}2^{3}2^{-5}} \\ \amp =\frac{2^{5+4-5} }{ ((2^2)^2)^{\frac{1}{2}}2^{3-5}} \\ \amp =\frac{2^{4} }{ 2^{2}2^{-2}} \\ \amp =2^{4 - (2 - 2) } \\ \amp =2^{4} \end{aligned}\)
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Checkpoint 2.6.17.

Checkpoint 2.6.18.

Determine the following for: \(x^2+ x^6+ x+ 3\)
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a) Determine the coefficient and the degree of each term.
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Term Coefficient Degree
\(x^2\)
\(x^6\)
\(x\)
\(3\)
b) The degree of the polynomial is ,
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the leading term is ,
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and the leading coefficient is .
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Answer 1.
\(1\)
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Answer 2.
\(2\)
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Answer 3.
\(1\)
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Answer 4.
\(6\)
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Answer 5.
\(1\)
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Answer 6.
\(1\)
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Answer 7.
\(3\)
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Answer 8.
\(0\)
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Answer 9.
\(6\)
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Answer 10.
\(1x^{6}\)
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Answer 11.
\(1\)
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Checkpoint 2.6.19.

The polynomial \(f(x) = \left(4-x\right)\mathopen{}\left(x+2\right)\mathopen{}\left(8x+3\right)^{2}\) has
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degree \(=\)
leading coefficient \(=\)
constant coefficient \(=\)
Answer 1.
\(4\)
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Answer 2.
\(-64\)
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Answer 3.
\(72\)
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Solution.
SOLUTION
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Since \(f(x) = \left(4-x\right)\mathopen{}\left(x+2\right)\mathopen{}\left(8x+3\right)^{2} = \left(-x+4\right)\mathopen{}\left(x+2\right)\mathopen{}\left(8x+3\right)\mathopen{}\left(8x+3\right)\text{,}\) by multiplying the highest degree terms in each factor we find the highest degree term to be \((-x)(x)(8 x)(8 x) = -64 x^{4}\text{,}\) and therefore the degree is \(4\) and the leading coefficent is \(-64\text{.}\) Similarly, by multiplying the constant terms from each factor of \(\left(-x+4\right)\mathopen{}\left(x+2\right)\mathopen{}\left(8x+3\right)\mathopen{}\left(8x+3\right)\text{,}\) we get the constant coefficient \((4)(2)(3)(3) = 72\text{.}\) Therefore, \(f(x) = -64 x^{4} + \cdots + 72\text{,}\) where we have omitted the degree 1, 2, and 3 terms.
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Checkpoint 2.6.20.

PTX:ERROR: WeBWorK problem Precalculus_Review_Materials/2_6_20.pg with seed 40 is either empty or failed to compile Use -a to halt with full PG and returned content
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