MPE Practice Problems for Math 147, 151, or 171

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Math Placement Exam

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Directions. The following are review problems for the MPE. We recommend you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "More Examples" to see similar examples. While there are no videos for the MPE problems, the similar examples do have videos explaining the solution.

Rationalize the denominator. $\displaystyle \frac{14}{3+\sqrt{2}}$

Answer
\[
\begin{align}
\left(\dfrac{14}{3+\sqrt{2}}\right) \left(\dfrac{3-\sqrt{2}}{3-\sqrt{2}}\right)&=\dfrac{42-14\sqrt{2}}{9-2}\\
&=\dfrac{7\left(6-2\sqrt{2}\right)}{7}\\
&=6-2\sqrt{2}
\end{align}
\]
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Section A.2 – Exponents and Radicals
Find the sum or difference as indicated, and write your answer in simplified form.\[\frac{x+2a-3}{x+a}-\frac{x+6}{2x}\]

Answer
\[\begin{align*}
\frac{x+2a-3}{x+a}-\frac{x+6}{2x} &=\left(\frac{x+2a-3}{x+a}\right) \left(\frac{2x}{2x}\right) - \left(\frac{x+6}{2x}\right)\left(\frac{x+a}{x+a}\right)\\
&=\frac{(x+2a-3)(2x)-(x+6)(x+a)}{(x+a)(2x)}\\
&=\frac{2x^2+4xa-6x-x^2-6x-xa-6a}{(x+a)(2x)}\\
&=\frac{x^2+3xa-12x-6a}{(x+a)(2x)}
\end{align*}\]

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Section A.4 – Rational Expressions
Factor and reduce to simplest form.\[\frac{6x^2+11xy-10y^2}{3x^2+10xy-8y^2}\]

Answer
\[ \dfrac{6x^2+11xy-10y^2}{3x^2+10xy-8y^2} = \dfrac{(3x-2y)(2x+5y)}{(3x-2y)(x+4y)}=\dfrac{2x+5y}{x+4y}\]
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Section A.4 – Rational Expressions
Simplify the following completely.\[\frac{\left(x^{-4}y^{2/5}\right)^{-3/4}}{x^{2/3}y^{-5/6}}\]

Answer
\[\begin{align*}
\frac{\left(x^{-4}y^{2/5}\right)^{-3/4}}{x^{2/3}y^{-5/6}}
&=\frac{x^{12/4}y^{-6/20}}{x^{2/3}y^{-5/6}}\\
&=x^{3-2/3}y^{-6/20-^-5/6}\\
&=x^{7/3}y^{32/60}\\
&=x^{7/3}y^{8/15}\\
\end{align*}\]
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Section A.2 – Exponents and Radicals
Solve the following equation: $5(x-7)-13(x-7)-6=0$.

Answer
\[\begin{align*}
5(x-7)-13(x-7)-6&=0\\
(x-7)(5-13)-6&=0\\
(x-7)(-8)-6&=0\\
-8x+56-6&=0\\
-8x+50&=0\\
-8x&=-50\\
x&=\dfrac{50}{8}
\end{align*}\]
Find the point $(x,y)$ which satisfies both equations. What is the value of $x+y$?\[\begin{align*} -2x+4y&=12\\ 3x-5y&=-3\end{align*}\]

Answer
Multiply $-2x+4y=12$ by $3$ and $3x-5y=-3$ by 2. This gives
\[\begin{align*}
-6x+12y&=36\\
6x-10y&=-6\\
\end{align*}\]
Add these two equations and we get $2y=30$ $\enspace \Longrightarrow \enspace$ $y=15$. Substituting $y=15$ into $3x-5y=-3$ gives $x=24$. So the point that satisfies both equations is $(24,15)$, and the value of $x+y$ is $24+15 =39$.
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Section 7.1&2 – Systems of Equations
Two investments are made, totaling $$10,000$. In one year, these investments yield $$650$ in simple interest. Part of the $$10,000$ is invested at $5\frac{1}{2}\%$, and the rest at $6\frac{3}{4}\%$. How much more money is invested at $6\frac{3}{4}\%$?

Answer
Let $x$ be the amount invested at $5\frac{1}{2}\%$ and $y$ be the amount invested at $6\frac{3}{4}\%$. The resulting system of equations is
\[\begin{align*}
x+y&=10,000\\
0.055x+0.0675y&=650
\end{align*}\]
Solve the system of equations to find $x$ and $y$.
$x=10,000-y$ $\enspace\Longrightarrow\enspace$ $0.055(10,000-y)+0.0675y=650$ $\enspace\Longrightarrow\enspace$ $550-0.055y+0.0675y=650$ $\enspace\Longrightarrow\enspace$ $0.0125y = 100$ $\enspace\Longrightarrow\enspace$ $y=8000$. Since $x+y=10,000$, we know that $x=2000$, and thus $$6,000$ more is invested at $6\frac{3}{4}\%$.
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Section 7.1&2 – Systems of Equations
Given the linear equation $2ax + 3by = 7c$, where $a$, $b$ and $c > 0$, if $x$ decreases by $10$ units, what is the corresponding change in $y$?

Answer
Since $2ax+3by=7c$ $\enspace\Longrightarrow\enspace$ $3by=7c-2ax$ $\enspace\Longrightarrow\enspace$ $y=\dfrac{7c-2ax}{3b}=\dfrac{-2a}{3b}x +\dfrac{7c}{3b}$. If $x$ decreases by $10$, then we know $y=\dfrac{-2a}{3b}(x-10)+\dfrac{7c}{3b}=\dfrac{-2ax}{3b}+\dfrac{20a}{3b}+\dfrac{7c}{3b}$, so $y$ will increase by $\dfrac{20a}{3b}$.
Line $A$ passes through the points $(2k+3, 4k−6)$ and $(−2, 16)$. Find the value of $k$ if line $A$ has a slope of $0$.

Answer
$\dfrac{4k-6-16}{2k+3+2}=0$ $\enspace\Longrightarrow\enspace$ $\dfrac{4k-22}{2k+5}=0$ $\enspace\Longrightarrow\enspace$ $4k-22=0$ $\enspace\Longrightarrow\enspace$ $k=\dfrac{22}{4}=5.5$
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Section 1.3 – Linear Equations in Two Variables
Jay wants to make a box, with no lid (or top), out of a $10'' \times 6''$ rectangular piece of cardboard. If Jay cuts squares with dimensions $x$ by $x$ out of each corner of the cardboard, and then folds up the corners to make an open box, find a function that represents:
1. The volume of the box.
2. The surface area of the box.
  Answer
  $V=(10-2x)(6-2x)x=4x^3-32x^2+60x$
  
  \[\begin{align*} SA&=2x(6-2x)+2x(10-2x)+(6-2x)(10-2x)\\&=12x-4x^2+20x-4x^2+60-32x+4x^2\\&=-4x^2+60\end{align*}\]
Solve for $x$ in the inequality $\dfrac{5x+2}{x-10}\geq 3$.

Answer
\[\begin{align*}
\dfrac{5x+2}{x-10}&\geq 3\\
\dfrac{5x+2}{x-10}-3&\geq 0\\
\dfrac{5x+2-3(x-10)}{x-10} &\geq 0\\
\dfrac{5x+2-3x+30}{x-10}&\geq 0\\
\dfrac{2x+32}{x-10}&\geq 0
\end{align*}\]
The expression $\dfrac{2x+32}{x-10}$ will be greater than or equal to zero on $(-\infty, -16]\cup(10,\infty)$.
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Section 2.7 – Nonlinear Inequalities
Find the domain of the function below.\[f(x)=\dfrac{\sqrt{x^2-3x-4}}{6x^2-54}\]

Answer
To find the domain of $ f(x)=\dfrac{\sqrt{x^2-3x-4}}{6x^2-54}$ we know $x^2-3x-4\geq 0$ and $6x^2-54 \neq 0$. $x^2-3x-4\geq 0$ for $(-\infty, -1]\cup[4,\infty)$ and $6x^2-54\neq 0$ for $(-\infty,-3)\cup(-3,3)\cup(3,\infty)$. So the domain of $f(x)$ is $ (-\infty,-3)\cup (-3,-1]\cup [4,\infty).$
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Section 1.4 – Functions
Find the domain of the function below. \[f(x)=\left\{\begin{array}{cc}\dfrac{2x^2+13}{x^2-1}, & x<0\\ \dfrac{5x-26}{x+2}, & x\geq 0\end{array} \right.\]

Answer
For the function $\dfrac{2x^2+13}{(x+1)(x-1)}$ on $ x<0$ the domain is $(-\infty,-1)\cup(-1,0)$. For the function $\dfrac{5x-26}{x+2}$ on $x\geq 0$ the domain is $[0,\infty)$. So for the function $f(x)$, the domain is $ (-\infty,-1)\cup (-1,\infty)$.
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Section 1.4 – Functions
Find the $x$-intercept(s) of the function $f(x)=\dfrac{6x^2-7x-5}{4x^2-12x-7}$, if any exist.

Answer
\[\begin{align*}
f(x)&=\dfrac{6x^2-7x-5}{4x^2-12x-7}\\
&=\dfrac{(2x+1)(3x-5)}{(2x+1)(2x-7)}\\
&=\dfrac{3x-5}{2x-7}
\end{align*}\]
The $x$-intercept is where $3x-5=0$ $\enspace\Longrightarrow\enspace$ $x=\dfrac{5}{3}$.
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Section 2.6 – Rational Functions
Find the vertical and horizontal asymptote(s) of the function $ f(x)=\dfrac{6x^2-7x-5}{4x^2-12x-7}$, if any exist.

Answer
\[\begin{align*}
f(x)&=\frac{6x^2-7x-5}{4x^2-12x-6}\\
&=\frac{(2x+1)(3x-5)}{(2x+1)(2x-7)}\\
&=\frac{3x-5}{2x-7}
\end{align*}\]
The vertical asymptote is where $2x-7=0$ so it is $x=\dfrac{7}{2}$.

The horizontal asymptote is at $y= \dfrac{6}{4}$ which reduces to $y= \dfrac{3}{2}$.
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Section 2.6 – Rational Functions
Find the $x$- and $y$-intercepts for the function $f(x)=x^3-9x$.

Answer
To find the $x$-intercepts solve $f(x)=0$ $\enspace \Longrightarrow\enspace $ $x(x+3)(x-3)=0$ $\enspace \Longrightarrow\enspace $ $x=0,\pm 3$.

To find the $y$-intercept, substitute $x=0$ $\enspace \Longrightarrow\enspace $ $0^3-9(0)=0$. Hence, the $y$-intercept is $y=0$.

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Section 2.2 – Polynomial Functions of Higher Degree
Find the domain of the following functions:
1. $f(x)=\sqrt{-x^2-4x+5}$
2. $g(t)=\ln(4t-3)$
3. $h(x)=\dfrac{1}{x^3+3x^2-x-3}$
  Answer
  We must have $−x^2−4x+5\geq 0$ for $f(x)$ to be defined. Thus, $(−x+1)(x+5)\geq 0$ which occurs on $[−5, 1]$. Thus, the domain of $f(x)$ is $[−5, 1]$.
  
  We must have $4t − 3 > 0$ for $g(t)$ to be defined. Thus, the domain for $g(t)$ is $\left(\dfrac{3}{4},\infty\right)$.
  
  We must have $x^3+3x^2-x-3\neq 0$ for $h(x)$ to be defined. \[\begin{align*} x^3+3x^2-x-3&\neq 0\\ x^2(x+3)-(x+3)&\neq 0\\ (x+3)\left(x^2-1\right) &\neq 0 \$x+3)(x-1)(x+1)&\neq 0 \end{align*}\] So \(x\neq \pm 1$ and $x\neq -3$. Thus, the domain of $h(x)$ is $(-\infty, -3)\cup (-3,-1) \cup (-1,1) \cup (1,\infty)$.
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  Section 1.4 – Functions
  
  Section 3.2 – Logarithmic Functions and Their Graphs
  
  Section 2.6 – Rational Functions
Simplify the expression $ \dfrac{2}{\sqrt{x^5}} \left(\sqrt[3]{4x}\right)$.

Answer
\[\begin{align*}
\left( \dfrac{2}{\sqrt{x^5}}\right) \left(\sqrt[3]{4x}\right) &= \left(2x^{-5/2} \right) (4x)^{1/3}\\
&=\left(2x^{-5/2}\right) \left(4^{1/3}\cdot x^{1/3}\right) \\
&=\left(2x^{-5/2}\right)\left( \left(2^2\right)^{1/3}x^{1/3}\right)\\
&=\left(2x^{-5/2}\right)\left(2^{2/3}x^{1/3}\right)\\
&=\dfrac{2^{5/3}}{x^{13/6}}
\end{align*}\]
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Section A.2 – Exponents and Radicals
If we begin with the graph of $f(x) = x^2$ and shift $f (x)$ 4 units to the right, shrink f (x) vertically by a factor of $\frac{1}{2}$, and then shift $f(x)$ upward 10 units, write the equation for the transformed graph.

Answer
The transformed graph is $g(x)=\dfrac{1}{2}(x-4)^2+10$.
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Section 1.6&7 – Parent Functions and Transformations
Solve the following equation for $x$.\[\log(x+2)+\log(x-1)=1\]

Answer
The domain of $\log(x+2)$ is $(-2,\infty)$. The domain of $\log(x-1)$ is $(1,\infty)$. So, the domain of $\log(x+2)+\log(x-1)$ is $(1,\infty)$. Now, use properties of logarithms to solve the equation.
\[\begin{align*}
\log(x+2)+\log(x-1)&=1\\
\log\left[ (x+2)(x-1)\right] &= 1\\
(x+2)(x-1)&=10^1\\
x^2+x-2&=10\\
x^2+x-12&=0\\
(x+4)(x-3)&=0\\
x&=-4,3
\end{align*}\]
Since $x=-4$ is not in the domain, $x=3$ is the only solution.

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Section 3.4 – Exponential and Logarithmic Equations
Factor the expression below completely. \[3x^2\left(4x^2+1\right)^8+64x^4\left(4x^2+1\right)^7\]

Answer
\[\begin{align*}
3x^2\left(4x^2+1\right)^8+64x^4\left(4x^2+1\right)^7 &= x^2\left(4x^2+1\right)^7 \left(3\left(4x^2+1\right)+64x^2\right)\\
&=x^2\left(4x^2+1\right)^7\left(12x^2+3+64x^2\right)\\
&=x^2\left(4x^2+1\right)^7\left(76x^2+3\right)
\end{align*}\]
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Section A.3 – Polynomials and Factoring
How far from the base of an $18$ foot tall pole must a person be standing if the angle of elevation from the ground to the pole is $41^\circ$?

Answer
Refer to the figure below. We know that $\tan (41^\circ)=\dfrac{18}{x}$. So, $x=\dfrac{18}{\tan 41^\circ}$ or $x=18\cot(41^\circ)$. So, the person must be $18\cot(41^\circ)$ feet from the base of the pole.

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Section 4.3 – Right Triangle Trigonometry
Find $f\circ g$ (also denoted $f(g(x))$) if $f(x)=\dfrac{x}{x+1}$ and $g(x)=\dfrac{2}{x}$. Simplify.

Answer
\[\begin{align*}
f\circ g &= \frac{ \frac{2}{x}}{\frac{2}{x}+1}\\
&= \frac{ \frac{2}{x}}{\frac{2}{x}+\frac{x}{x}}\\
&=\frac{ \frac{2}{x}}{\frac{2+x}{x}}\\
&=\frac{2}{x}\cdot \frac{x}{2+x} \\
&=\frac{2}{2+x}
\end{align*}\]
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Section 1.8 – Combinations of Functions

Section A.4 – Rational Expressions
Perform the indicated operations and simplify. \[\frac{8}{x+1}-\left(\frac{y}{z+2} \div \frac{y-4}{w}\right)\]

Answer
\[\begin{align*}
\frac{8}{x+1}-\left( \frac{y}{z+2} \div \frac{y-4}{w}\right) &= \frac{8}{x+1}-\left( \frac{y}{z+2} \cdot \frac{w}{y-4}\right)\\
&=\frac{8}{x+1} - \left( \frac{yw}{(z+2)(y-4)}\right) \\
&=\frac{ 8(z+2)(y-4)-yw(x+1)}{(x+1)(z+2)(y-4)}\\
&=\frac{8zy-32z+16y-64-ywx-yw}{(x+1)(z+2)(y-4)}
\end{align*}\]
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Section A.4 – Rational Expressions
Solve the equation $e^{2x}-2e^x-3=0$ for $x$.

Answer
First, factor the equation: $e^{2x}-2e^x-3=(e^x)^2-2e^x-3=(e^x-3)(e^x+1)$. Solving $(e^x-3)(e^x+1)=0$ $\enspace \Longrightarrow\enspace $ $e^x-3=0$ or $e^x+1=0$ $\enspace \Longrightarrow\enspace $ $e^x=3$ or $e^x=-1$. But $e^x$ will never be negative, so the only solution is $e^x=3$ $\enspace \Longrightarrow\enspace $ $x=\ln(3)$.
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Section 3.4 – Exponential and Logarithmic Equations
Find the equation of the line passing through the point $(5,1)$ with a slope of 7. Use the equation you find to determine the value of $y$ when $x=-4.$

Answer
Using the point-slope equation, we get $y-1=7(x-5)$ $\enspace \Longrightarrow\enspace $ $y=7x-35+1$ $\enspace \Longrightarrow\enspace $ $y=7x-34$. Using this to find $y$ when $x=-4$ gives $y=7(-4)-34$ $\enspace \Longrightarrow\enspace $ $y=-62$.
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Section 1.3 – Linear Equations in Two Variables
If $f(x)=\sqrt{x+4}$, find and simplify $\dfrac{f(2+h)-f(2)}{h}$.

Answer
\[\begin{align*}
\dfrac{f(2+h)-f(2)}{h} &=\dfrac{\sqrt{2+h+4}-\sqrt{2+4}}{h}\\
&=\left(\dfrac{\sqrt{6+h}-\sqrt{6}}{h}\right) \cdot \left( \dfrac{\sqrt{6+h}+\sqrt{6}}{\sqrt{6+h}+\sqrt{6}}\right)\\
&=\dfrac{6+h-6}{h\left(\sqrt{6+h}+\sqrt{6}\right)}\\
&=\dfrac{1}{\sqrt{6+h}+\sqrt{6}}
\end{align*}\]
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Section 1.4 – Difference Quotient
Simplify $\dfrac{ \left(x^2y^4\right)^5\left( x^3y\right)^{-3}}{xy}$.

Answer
\[\begin{align*}
\dfrac{ \left(x^2y^4\right)^5\left( x^3y\right)^{-3}}{xy}
&=\dfrac{x^{10}y^{20}x^{-9}y^{-3}}{xy}\\
&=\dfrac{xy^{17}}{xy}\\
&=y^{16}
\end{align*}\]
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Section A.2 – Exponents and Radicals
Simplify $\sqrt[3]{\left(a^3b\right)}\sqrt[3]{64a^4b^2}$.

Answer
$\sqrt[3]{\left(a^3b\right)}\sqrt[3]{64a^4b^2}=\sqrt[3]{64a^7b^3}=4a^2b\left(\sqrt[3]{a}\right)$
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Section A.2 – Exponents and Radicals
Perform the operations indicated and simplify.\[\frac{x^2}{x^2-x-2}-\frac{4}{x^2+x-6}+\frac{x}{x^2+4x+3}\]

Answer
\[\begin{align*}
\frac{x^2}{x^2-x-2}&-\frac{4}{x^2+x-6}+\frac{x}{x^2+4x+3}\\
&=\frac{x^2}{(x-2)(x+1)}-\frac{4}{(x+3)(x-2)}+\frac{x}{(x+1)(x+3)}\\
&=\left(\frac{x+3}{x+3}\right)\left(\frac{x^2}{(x-2)(x+1)}\right) -\left(\frac{x+1}{x+1}\right) \left(\frac{4}{(x+3)(x-2)}\right)+\left(\frac{x-2}{x-2}\right)\left(\frac{x}{(x+1)(x+3)}\right)\\
&= \frac{x^3+3x^2-4x-4+x^2-2x}{(x+3)(x-2)(x+1)}\\
&=\frac{x^3+4x^2-6x-4}{(x+3)(x-2)(x+1)}
\end{align*}\]
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Section A.4 – Rational Expressions
Find all zeros, horizontal, and vertical asymptotes for $f(x)=\dfrac{3x^2-14x-5}{4x^2-17x-15}$.

Answer
Factor and simplify: $f(x)=\dfrac{3x^2-14x-5}{4x^2-17x-15}=\dfrac{(3x+1)(x-5)}{(4x+3)(x-5)}=\dfrac{3x+1}{4x+3}$. This yields a zero of $x=-\dfrac{1}{3}$, a vertical asymptote of $x=-\dfrac{3}{4}$, and a horizontal asymptote of $y=\dfrac{3}{4}$.
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Section 2.6 – Rational Functions
If $\theta$ is in Quadrant II, and $\sin(\theta) = \dfrac{1}{7}$, what is $\cos(\theta)$?

Answer
Using the identity $\cos^2 \theta + \sin^2 \theta = 1$, we find that $\cos^2 \theta = 1-\sin^2\theta=1-\dfrac{1}{49}=\dfrac{48}{49}$. This gives us $\cos \theta= \pm \sqrt{\dfrac{48}{49}}=\pm \dfrac{\sqrt{48}}{7} = \pm \dfrac{4\sqrt{3}}{7}$. Since $\theta$ is in quadrant II, $\cos \theta <0$, hence $\cos \theta = -\dfrac{4\sqrt{3}}{7}$.
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Section 4.4 – Trigonometric Functions of Any Angle
Use the properties of logarithms to expand the expression $ \ln \left( \dfrac{\sqrt{x}y^5}{(z+1)^4}\right)$.

Answer
We know $\ln(ab)=\ln a + \ln b$, $ \ln \left(\dfrac{a}{b}\right)=\ln a - \ln b$, and $ \ln a^b =b \ln a$. Using these properties, we obtain
\[\begin{align*}
\ln \left( \frac{ \sqrt{x}y^5}{(z+1)^4} \right) &= \ln \left( \sqrt{x}y^5\right) - \ln (z+1)^4\\
&=\ln \left(x^{1/2}y^5\right)-4\ln(z+1)\\
&=\ln x^{1/2}+\ln y^5 -4\ln(z+1)\\
&=\dfrac{1}{2}\ln x+5\ln y -4\ln(z+1)
\end{align*}\]
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Section 3.3 – Properties of Logarithms
Evaluate $ \sec \dfrac{2\pi}{3} - \tan \dfrac{\pi}{6}$.

Answer
$\sec \dfrac{2\pi}{3}-\tan \dfrac{\pi}{6} = -2 - \dfrac{1}{\sqrt{3}}$
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Section 4.2 – Trigonometric Functions: The Unit Circle
If we begin with a rectangle with length 5 inches and width 4 inches, then increase the length by $8 \%$, what is the change in area?

Answer
We begin with a rectangle of area $20$ in$^2$. If we increase the length by $8 \%$, then our new length is $5.4$ inches. This gives a new area of $21.6$ in$^2$. This gives a total change in area of $1.6$ in$^2$.
Evaluate $ f(2)-f(-3)$ given $ f(x)=\left\{ \begin{array}{ll} x^3+1, & x>1\\2x^2-3,&x\leq 1\end{array}\right.$

Answer
$f(2)=(2)^3+1=9$ and $f(-3)=2(-3)^2-3=18-3=15$, so $f(2)-f(-3)=9-15=-6$.
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Section 1.4 – Functions
Simplify the expression $\dfrac{\cos^2 (\theta)}{1+\sin(\theta)}$.

Answer
Use the identity $\cos^2\theta = 1-\sin^2 \theta$ to get
\[\begin{align*}
\frac{\cos^2(\theta)}{1+\sin(\theta)} &= \frac{1-\sin^2\theta}{1+\sin \theta}\\
&=\frac{(1+\sin \theta)(1-\sin \theta)}{1+\sin\theta}\\
&=1-\sin \theta
\end{align*}\]
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Section 5.2 – Verifying Trigonometric Identities
Evaluate $\log_4\left(\dfrac{1}{\sqrt[3]{16}}\right)$.

Answer
\[\begin{align*}
\log_4 \left(\dfrac{1}{\sqrt[3]{16}}\right) &=\log_4\left( \frac{1}{\sqrt[3]{4^2}}\right)\\
&=\log_4 \left( \dfrac{1}{4^{2/3}}\right)\\
&=\log_4 4^{-2/3} \\
&=-\dfrac{2}{3}\\
\end{align*}\]
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Section 3.2 – Logarithmic Functions and Their Graphs
Simplify $\dfrac{\frac{1}{a}-b}{\frac{1}{b^3}+a}$.

Answer
$\dfrac{\frac{1}{a}-b}{\frac{1}{b^3}+a} =
\dfrac{\left(\frac{1}{a}-b\right)\left(ab^3\right)}{\left(\frac{1}{b^3}+a\right)\left(ab^3\right)}
=\dfrac{b^3-ab^4}{a+a^2b^3}$
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Section A.4 – Rational Expressions
A bacteria contains $1200$ bacteria and doubles every day. How many hours will it take the culture to reach 10000 bacteria?

Answer
Use the exponential model $y(t)=y_0e^{kt}$, where $y(t)$ is the size of the population at time $t$ and $y_0$ is the initial size of the population. We know that $y_0=1200$, hence $y(t)=1200e^{kt}$. Since the population doubles everyday, we know
\[\begin{align*}
2400&=1200e^{k\cdot 24}\\
2&=e^{24k}\\
\ln 2&= 24 k\\
k&=\dfrac{1}{24}\ln 2\\
k&= \ln 2^{1/24}\\
\end{align*}\]
Thus, $y(t)=1200e^{t\ln 2^{1/24}} = 1200 e^{\ln 2^{t/24}}=1200\cdot 2^{t/24}$ (note $t$ is in hours). Now solve for $t$ when $y(t)=10000$
\[\begin{align*}
10000&=1200\cdot 2^{t/24}\\
\dfrac{10000}{1200}&=2^{t/24}\\
\dfrac{25}{3}&=2^{t/24}\\
\log_2\left(\dfrac{25}{3}\right) &= \dfrac{t}{24}\\
t&=24\log_2 \left(\dfrac{25}{3}\right)\\
\end{align*}\]
So, it will take $24 \log_2 \left( \dfrac{25}{3}\right)$ hours for the culture to reach $10,000$ bacteria.
More Examples
Follow the link below to see more examples.
Section 3.5 – Exponential and Logarithmic Models