Section 2.6 Exercises
These are the exercises in Module 2.
Checkpoint 2.6.2.
Checkpoint 2.6.3.
Enter Yes or No in each answer space below to indicate whether the corresponding equation defines \(y\) as a function of \(x\text{.}\)
1. \(x^2y+y= 9\)
2. \(4 x = y^2\)
3. \(4 + x = y^3\)
4. \(x + 4 = y^2\)
Checkpoint 2.6.4.
The domain of the function
\(f(x)=\sqrt{x(x-14)}\)
is
Write the answer in interval notation.
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.
Answer.
\(\left(-\infty ,0\right]\cup \left[14,\infty \right)\)
Checkpoint 2.6.5.
One of the solutions of \((5x+2)(x+3)=0\) is:
- \(\displaystyle \frac{5}{2} \)
- \(\displaystyle -\frac{2}{5} \)
- \(\displaystyle \frac{2}{5} \)
- \(\displaystyle 3 \)
- None of the above
Answer.
\(\text{Choice 2}\)
Checkpoint 2.6.6.
Checkpoint 2.6.7.
Find all values of \(x\) that satisfy the equation
\(9(x+8)+2=-8(x-7)-4\)
List the values below, seperated by commas.
Values of \(x\) =
Answer.
\(\frac{-22}{17}\)
Checkpoint 2.6.8.
Find all real zeros of \(f(x)= x^2 - 6 x - 7\text{.}\)
Zeros are \(x =\) .
Note: If there is more than one answer enter them separated by commas.
Answer.
\(-1, 7\)
Checkpoint 2.6.9.
Select the expression that is equivalent to
\(\frac{x-1}{5x}\cdot\frac{x}{x^2-1}\)
- \(\displaystyle \frac{-1}{5(x-1)} \)
- \(\displaystyle \frac{1}{5(x+1)}, x \neq 0, x \neq 1 \)
- \(\displaystyle \frac{-1}{5(x-1)}, x \neq 0, x \neq 1 \)
- \(\displaystyle \frac{1}{5(x+1)} \)
- \(\displaystyle \frac{-1}{5(x-1)}, x \neq 0 \)
- \(\displaystyle \frac{1}{5(x+1)}, x \neq 0 \)
- None of the above
Answer.
\(\text{Choice 2}\)
Checkpoint 2.6.10.
Write \(\frac{m}{2}+\frac{m}{3}\) as a single fraction.
- \(\displaystyle \frac{2m}{6} \)
- \(\displaystyle \frac{m^2}{6} \)
- \(\displaystyle \frac{5m}{5} \)
- \(\displaystyle \frac{5m}{6} \)
- \(\displaystyle \frac{m}{5} \)
- \(\displaystyle \frac{2m}{5} \)
- None of the above
Answer.
\(\text{Choice 4}\)
Checkpoint 2.6.11.
Select the expression that is equivalent to
\(\frac{x-2}{5x}\cdot\frac{x}{x^2-4}\)
- \(\displaystyle \frac{-1}{5(x-2)}, x \neq 0 \)
- \(\displaystyle \frac{1}{5(x+2)}, x \neq 0 \)
- \(\displaystyle \frac{1}{5(x+2)} \)
- \(\displaystyle \frac{1}{5(x+2)}, x \neq 0, x \neq 2 \)
- \(\displaystyle \frac{-1}{5(x-2)}, x \neq 0, x \neq 2 \)
- \(\displaystyle \frac{-1}{5(x-2)} \)
- None of the above
Answer.
\(\text{Choice 4}\)
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.
\(\displaystyle \left( \frac{27}{8} \right)^{-1/3}\) =
Answer.
\({\frac{2}{3}}\)
Checkpoint 2.6.14.
Checkpoint 2.6.15.
The expression
\(\left(y^{\frac{3}{6}}\right)^{\frac{2}{5}}\)
equals \(y^r\) where \(r\text{,}\) the exponent of \(y\text{,}\) is:
Answer.
\({\frac{1}{5}}\)
Checkpoint 2.6.16.
The expression
\begin{equation*}
\frac{2^52^42^{-5}}{\sqrt{4^2}2^32^{-5}}
\end{equation*}
equals \(2^n\) where \(n\) is:
Checkpoint 2.6.17.
Checkpoint 2.6.18.
Determine the following for: \(x^2+ x^6+ x+ 3\)
a) Determine the coefficient and the degree of each term.
Term | Coefficient | Degree |
\(x^2\) | ||
\(x^6\) | ||
\(x\) | ||
\(3\) |
b) The degree of the polynomial is ,
the leading term is ,
and the leading coefficient is .
Checkpoint 2.6.19.
The polynomial \(f(x) = \left(4-x\right)\!\left(x+2\right)\!\left(8x+2\right)^{2}\) has
degree | \(=\) | |
leading coefficient | \(=\) | |
constant coefficient | \(=\) |
Answer 1.
Answer 2.
Answer 3.
Solution.
\(4\)
\(-64\)
\(32\)
SOLUTION
Since \(f(x) = \left(4-x\right)\!\left(x+2\right)\!\left(8x+2\right)^{2} = \left(-x+4\right)\!\left(x+2\right)\!\left(8x+2\right)\!\left(8x+2\right)\text{,}\) 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 \(\left(-x+4\right)\!\left(x+2\right)\!\left(8x+2\right)\!\left(8x+2\right)\text{,}\) we get the constant coefficient \((4)(2)(2)(2) = 32\text{.}\) Therefore, \(f(x) = -64 x^{4} + \cdots + 32\text{,}\) where we have omitted the degree 1, 2, and 3 terms.
Checkpoint 2.6.20.
Given that \(f(x)\) is a cubic function with zeros at \(-6\text{,}\) \(-1\text{,}\) and \(7\text{,}\) find an equation for \(f(x)\) given that \(f(-8)=-2\text{.}\)
Answer.
\(\left({\frac{1}{105}}\right)\!\left(x+6\right)\!\left(x+1\right)\!\left(x-7\right)\)