MATH 112 – Final Exam Practice | A minimal Hugo website

How to use this page
Try each problem first, then expand Hint or Solution as needed. (this is a draft. Currently in the process of checking and reviewing the solutions are correct. I hope It Helps you in your study.)

Formulas Provided on the Exam {: #formula-sheet }

Quadratics

Quadratic Formula	Vertex of Parabola
$x = \dfrac{-b \pm \sqrt{b^{2}-4ac}}{2a}$	$x = -\dfrac{b}{2a}$

Factored Forms

Form
$y = a(x-p)(x-q)$
$y = a(x-h)^2 + k$
$y = a(x-r_1)(x-r_2)$

Exponentials & Logs

Exponential Growth / Decay	Logs Conversion
$A = P\!\bigl(1+\tfrac{r}{n}\bigr)^{nt}$	$\log_b x = \dfrac{\ln x}{\ln b}$
$A = P e^{rt}$

Compound Interest

Continuous Compounding
$A = P e^{rt}$

Problems

1 Sales‑Tax Expression

In a certain city, sales tax is 9 %. Write an expression for the total cost of an item priced $x$ dollars after tax is added.

A. $C(x)=1.09x$

B. $C(x)=0.09x$

C. $C(x)=1.9x$

D. $C(x)=9x$

E. $C(x)=0.91x$

Hint

Add 9 % of the price to the price to the total cost.

Solution

9 % of $x$ is $ \frac{9}{100}x = 0.09x$; total cost $C(x) = x+0.09x = 1.09x$. Answer A.

2 Which Graph is a Function?

Graph A

Hint

Use the vertical line test.

Solution

Graph A passes the vertical‑line test, so only Graph A is a function.

3 Domain of $R(x)=\sqrt{2-5x}$

A. $[0.4,\infty)$

B. $(-\infty,0.4]$

C. $(0.4,\infty)$

D. $(-\infty,0.4)$

E. $[0,\infty)$

Hint

input of a Square roots cannot be negative.

Solution

The inside of the squarer root cannot admit negative values. So we require $$2-5x \geq 0.$$ solve for x. $$\begin{align*} 2 & \geq 5x \\ \frac{2}{5} &\geq x \end{align*}$$ Hence the domain of $ x $ must be $ (-\infty, \frac{2}{5}) $. Which can be written as $ (-\infty,0.4] $. Answer B.

4 Intervals of Increase

Piece‑wise curve for Q 4

A. $(-\infty,3)$

B. $(-\infty,0)$

C. $(0,3)$

D. $[3,\infty)$

E. $(0,3)\cup(3,\infty)$

Solution

From the sketch the function rises for all $x \geq 3 $ so the answer is $ [3, \infty) $.

5 Starbucks — Slope & Interpretation

Year	Locations
2012	18 066
2014	21 366
2015	23 016

What is the slope of the line that passes through these points, and what does it mean in practical terms?

A. $m = 3300$. The number of Starbucks locations worldwide increases by 3300 per year.

B. $m = 3300$. The number of locations increases by 3300 over 3 years.

C. $m = 4950$. The number of locations increases by 4950 per year.

D. $m = 1650$. The number of locations increases by 1650 over 3 years.

E. $m = 1650$. The number of locations worldwide increases by 1650 per year.

Solution

Slope

\[ m=\frac{23\,016-18\,066}{2015-2012}=\frac{4965}{3}=1650. \]

Thus Starbucks was adding about 1650 locations each year over the period shown.
Answer E.

6 Linear Model for Starbucks Locations

Which linear function best models the number of Starbucks locations $S(t)$ as a function of time $t$ in years since 2012?

A. $S(t)=1650t+18\,066$

B. $S(t)=3300t+18\,066$

C. $S(t)=4950t+18\,066$

D. $S(t)=1650t+21\,366$

E. $S(t)=3300t+21\,366$

Solution – TODO

at $t=0$ (2012), $S(0)=18\,066$. Slope $m=1650$ locations per year. \[ S(t)=1650t+18\,066\] So the answer is $S(t)=1650t+18\,066$. Answer A.

7 Evaluate the Piecewise Function

\[ f(x)= \begin{cases} 2x+5 & x<-2\\[2pt] x-1 & -2\le x\le1\\[2pt] \tfrac13x+4 & x>1 \end{cases} \]

Evaluate $f(-3)$.

A. 3 B. –1 C. –4 D. 1 E. undefined

Hint

In which branch does $ x= -3 $ fall? Use corresponding formula to compute $f(-3)$.

Solution

Since $-3<-2$, use the first branch: \[ f(-3)=2(-3)+5=1.\] So the answer is $f(-3)=1$. Answer D.

8 (*) Transformations of $g(x)$

If $(2,5)$ lies on $y=g(x)$, which point must lie on
$y=\tfrac15$, $2g\!\bigl(\tfrac{1}{5}X\bigr)+3$?

A. $\bigl(\tfrac25,13\bigr)$
B. $(4,4)$
C. $(10,13)$
D. $\bigl(\tfrac{11}{10},2\bigr)$
E. $(4,13)$

Hint

Identify input and output of the function and the transformations.

Solution

1. To send the old input $2$ through $x/5$, set $\frac{x}{5}=2$ ⇒ $x=10$. 2. The original output is $g(2)=5$. Then \[ y = 2\cdot5 + 3 = 13. \] Hence the image is \[ \boxed{(10,13)}. \]

---

9 Fuel‑Efficiency Scaling

$d=f(x)$ gives distance (mi) for $x$ gallons of gas. Point $(10,250)$ is on the graph.
In five years the same car travels twice as far per gallon. Which point lies on the new graph?

A. $(20,250)$

B. $(20,500)$

C. $(10,125)$

D. $(10,500)$

E. $(20,125)$

Hint

Identify input and output of the function. Which point is scaled by a factor of 2?

Hint

Originally $10$ gal ⇒ $250$ mi. Doubling fuel efficiency means the same $10$ gal ⇒ $2\times250=500$ mi. \[ \boxed{(10,500)}. \]

10. (***) Composition of Rational Functions

Consider the functions

\[ f(x) = \frac{x - 1}{x} \qquad\text{and}\qquad g(x) = \frac{x - 2}{x + 5}. \]

Find $(f\circ g)(x)$.

A. $\displaystyle (f\circ g)(x) = -\frac{7}{x^{2} + 5x}$

B. $\displaystyle (f\circ g)(x) = \frac{x^{2} - 3x + 2}{x^{2} + 5x}$

C. $\displaystyle (f\circ g)(x) = -\frac{x - 1}{6x - 1}$

D. $\displaystyle (f\circ g)(x) = -\frac{x - 1}{x^{2} + 5x}$

E. $\displaystyle (f\circ g)(x) = -\frac{7}{x - 2}$

Hint

Identify the inner and outer functions: \[ (f\circ g)(x) = f\bigl(g(x)\bigr). \] First compute $g(x)$, then substitute that result into $f$.

Hint

Identify the inner and outer functions: \[ (f\circ g)(x) = f\bigl(g(x)\bigr). \] First compute $g(x)$, then substitute into $f$.

Solution

\[ (f\circ g)(x) = \frac{g(x)-1}{g(x)} \quad\text{with}\quad g(x)=\frac{x-2}{x+5}. \] \[ \begin{aligned} \frac{g(x)-1}{g(x)} &= \frac{\dfrac{x-2}{x+5}-1} {\dfrac{x-2}{x+5}}\\ &= \frac{\dfrac{x-2}{\cancel{x+5}} - \dfrac{x+5}{\cancel{x+5}}} {\dfrac{x-2}{\cancel{x+5}}}\\ &= \frac{x-2-(x+5)}{x-2}\\ &= \frac{-7}{x-2} \end{aligned} \]

11 (*) Coupon Order Matters

Kohl’s allows customers to use both a “$10 off” coupon and a “30 % off” coupon on the same transaction, but the store stipulates that the “dollar off” coupon must be used first.

Define

\[ f(x) = x - 10 \quad\text{and}\quad g(x) = 0.70x. \]

Which of the following functions represents the final price after applying first the $\$10$‑off coupon and then the 30 %‑off coupon?

A. $(g\circ f)(x)$
B. $(f\circ g)(x)$
C. $(f\cdot g)(x)$
D. $(f+g)(x)$
E. $(g-f)(x)$

Hint

Since the $\$10$‑off must be applied first, feed the original price $x$ into $f$, then take that result and feed it into $g$.

Solution

We want to subtract $\$10$, then take 70 % of the remainder. That is \[ (g\circ f)(x) = g\bigl(f(x)\bigr) = 0.70\bigl(x - 10\bigr). \] None of the other orders or operations matches this. Hence the correct answer is **A. $(g\circ f)(x)$**.

---

12 (*) Inverse Function

Find $f^{-1}(x)$ for $f(x)=\dfrac{x-1}{2x+3}$.

A. $f^{-1}(x) = \dfrac{2x+3}{x-1}$

B. $f^{-1}(x) = \dfrac{-3x-1}{2x-1}$

C. $f^{-1}(x) = \dfrac{x+4}{2x+3}$

D. $f^{-1}(x) = \dfrac{-x+1}{x+2}$

E. None of these

Hint

solve for x.

Solution

solve for x. \[ y = \dfrac{x-1}{2x+3} \implies y(2x+3) = x-1 \implies 2xy + 3y = x - 1\] Put all $x$ terms on one the same side and factor out $x$: \[ x - 2xy = 3y + 1 \implies x(1-2y) = 3y + 1 \implies x = \dfrac{3y+1}{1-2y} \] Now switch $x$ and $y$: \[ f^{-1}(x) = \dfrac{3x+1}{1-2x}\] So the answer is $f^{-1}(x) = \dfrac{3x+1}{1-2x}$. Answer B.

13 (*) Evaluate $f^{-1}(2)$ from a Graph

Graph for inverse‑value question

A. 0 B. 2 C. –2 D. Undefined E. None of these

Hint

$g(3) =2$ means when the input is 3, the output is 2. The inverse function $g^{-1}(2)$ asks: what is the input that gives an output of 2?

Solution

The graph shows that $g(-2)=2$, so the inverse function $g^{-1}(2)=-2$. Thus the answer is B.

14 Find a Quadratic from a Graph

Graph of quadratic

A. $y=(x+1)(x-2)$

B. $y=-(x+1)(x-2)$

C. $y=(x-2)(x-1)$

D. $y=-(x-2)(x-1)$

E. $y=-(x+1)(x-2)-$

Hint

Identify zeros & orientation.

Solution

Observe the $y=0$ when $x=-1$ or $x=2$. Thus the factored form is \[ y = A(x+1)(x-2).\] Now we evaluate the formula above at the point (0,2) to find $A$: \[ 2 = A(0+1)(0-2) \implies 2 = -2A \implies A = -1.\] So the equation is \[ y = -(x+1)(x-2).\] Thus the answer is $y=-(x+1)(x-2)$. Answer B.

15 Time for Projectile to Land

A stone’s height (m) after $t$ s:
$h(t)=-4.9t^{2}+49t+277.4$. When does it hit the ground?

A. < 9 s

B. 9–11 s

C. 11–13 s

D. 13–15 s

E. > 15 s

Hint

Solve h(t) = 0. Use quadratic formula.

Solution: Calculator version

There are two ways to solve this problem with the claculator.

Version 1: program > quad > “asks you for a? b? c? > gives you the roots.

Version 2:

Press Y=
Enter your quadratic, e.g.
```
Y1 = A·X² + B·X + C
```
Press GRAPH
Press 2nd → CALC
Select 2: Zero
Left Bound? Move the cursor to just left of the root and press ENTER
Right Bound? Move the cursor to just right of the root and press ENTER
Guess? Just press ENTER — the calculator will display the root.

Solution

Set $h(t)=0$: \[ 0=-4.9t^{2}+49t+277.4\] Use the quadratic formula: \[ t=\dfrac{-49\pm\sqrt{49^{2}-4(-4.9)(277.4)}}{2(-4.9)}\] \[ t=\dfrac{-49\pm\sqrt{2401+5443.76}}{-9.8}\] \[ t=\dfrac{-49\pm\sqrt{7844.76}}{-9.8}\] \[ t=\dfrac{-49\pm88.6}{-9.8}\] \[ t=\dfrac{-49+88.6}{-9.8}\approx-4.02\text{ (not valid)}\] \[ t=\dfrac{-49-88.6}{-9.8}\approx14.05\text{ (valid)}\] So the stone hits the ground at about $t=14$ seconds. Thus the answer is D

16 (*) Profit Maximization

Demand: $p(x)=50-\dfrac{1}{4000}x$. Fixed cost $25 000; variable $3 per pair.
How many sunglasses maximize profit?

A. 100 000

B. 87 000

C. 94 000

D. 200 000 E. 188 000

Hint.

Combine formulas to obtain an equation for the profit: Revenue = demand x price Profit = Revenue - cost. Obtain a parabola. find vertex.

Solution

Revenue: \[ R(x)=p(x)x=(50-\dfrac{1}{4000}x)x=50x-\dfrac{1}{4000}x^{2}\] Cost: \[ C(x)=25000+3x\] Profit: \[ P(x)=R(x)-C(x)=\left(50x-\dfrac{1}{4000}x^{2}\right)-(25000+3x)\] \[ P(x)=-\dfrac{1}{4000}x^{2}+47x-25000\] This is a downward-opening parabola. The maximum occurs at the vertex: \[ x=-\dfrac{b}{2a}=-\dfrac{47}{2(-\dfrac{1}{4000})}=47\cdot2000=94000.\] Thus the maximum profit occurs at $x=94\,000$ pairs. So the answer is C.

17 Facebook Users Model

The number of monthly active Facebook users worldwide (in millions) between 2009 and 2017 is modeled by

\[ f(t)=2.12\,t^3 \;-\;27.43\,t^2 \;+\;301.48\,t \;+\;180.4, \]

where $t$ is years since 2009. Use this model to answer the next question.

17. Evaluate $f(1)$ and give a practical interpretation.

A. $f(1)=456.57$. In the year 2010, there were an average of 456.57 Facebook users added each month.
B. $f(1)=456.57$. In the year 2010, there were 456.57 monthly Facebook users.
C. $f(1)=456.57$. In the year 2010, there were 456.57 million monthly Facebook users.
D. $f(1)=456.57$. In the year 2010, there were an average of 456.57 million Facebook users added each month.
E. $f(1)=456.57$. To reach 1 million monthly Facebook users, it took 456.57 months.

Hint

Substitute $t=1$ into the model. Remember that $f(t)$ is in **millions** of users.

Solution

\[ f(1) =2.12(1)^3 -27.43(1)^2 +301.48(1) +180.4 =456.57. \] Since $f$ measures millions of users, $456.57$ means **456.57 million monthly active users** in 2010. **Answer C.**

18 (***) When Did Users Reach 1 Billion?

According to this model, in what year did the number of monthly active Facebook users reach 1 000 million?

A. 2011 B. 2012 C. 2013 D. 2014 E. 2015

Hint

Solve $f(t)=1000$ (millions) for $t$, then add that to 2009. or alternative use calculator and intersect f(t) with y=1000.

Solution

Use calculator. We obtain

19 Factor $R(x)$ Given a Root

Graph

\[ R(x)=2x^3 + 5x^2 - 6x - 1 \]

on your calculator. Given that $(x-1)$ is a factor, what are the other linear factors?

A. $(2x+3)$ and $(x+2)$

B. $(2x-3)$ and $(x-2)$

C. $(3x+1)$ and $(x-3)$

D. $(3x-1)$ and $(x+3)$

E. $(2x+1)$ and $(x+6)$

Hint

Divide $R(x)$ by $(x-1)$ to obtain a quadratic, try graphing such quadratic to find its roots.

Hint

find roots of $$ \frac{2x^3 + 5x^2 - 6x - 1}{x-1}.$$

Solution

Graph $$R(x)= \frac{2x^3 + 5x^2 - 6x - 1}{x-1}$$ and find its intersection with $$y=0$$.

The roots are $x=-2$ and $x=\frac{-3}{2}$. verify your answer: R(x) = (x-1)(2x+3)(x+2)

\[ \text{Verify: } R(-2) = 0 \text{ and } R\left(-\frac{3}{2}\right) = 0 \]

20 End Behavior of $ f(t)=\dfrac{6t}{t + 1.2} $

As $t$ gets very large, what does $f(t)$ approach?

A. $f(t)\to\infty$

B. $f(t)\to1.2$

C. $f(t)\to3.8$

D. $f(t)\to6$

E. None of these

Hint

Take a look at the leading terms.

Solution

The leading terms are $\(6t$ and $t$). so as $t\to\infty$, \[ \lim_{t\to\infty}\frac{6t}{t+1.2} =\lim_{t\to\infty}\frac{6}{1 + \tfrac{1.2}{t}} =6. \]

\[ \boxed{6}. \]

Answer D.

21 (***) Evaluate $h^{-1}(5)$ and Interpret

The height (ft) of a dwarf evergreen tree after $t$ years is

$$ h(t) = \frac{6t}{1.2 + t}. $$

Find $h^{-1}(5)$, i.e.\ the time $t$ when the tree reaches 5 ft.

A. $h^{-1}(5)=4.84$. When a tree is 5 ft tall, 4.84 years have passed.

B. $h^{-1}(5)=1.2$. When a tree is 5 ft tall, 1.2 years have passed.

C. $h^{-1}(5)=6$. When a tree is 5 ft tall, 6 years have passed.

D. $h^{-1}(5)=4.84$. After 5 years, the height is 4.84 ft.

E. $h^{-1}(5)=6$. After 5 years, the height is 6 ft.

Hint

Solve $h(t) = 5$.

Solution

We want $t$ such that \[ \frac{6t}{1.2 + t} = 5. \] Multiply both sides by $1.2+t$, distribute, and isolate $t$: $$ \begin{aligned} 6t &= 5(1.2 + t) &&\bigl|\times(1.2+t)\\ 6t &= 6 + 5t &&\bigl|\,-5t\\ 6t - 5t &= 6\\ t &= 6. \end{aligned} $$

Thus $h^{-1}(5)=6$, meaning when the tree reaches 5 ft tall, 6 years have passed.

$$ \boxed{6}. $$

Choice C.

22 (***) Horizontal Asymptote of $\displaystyle p(x)=\frac{2-3x}{x+1}$

A. $y=-3$

B. $y=2$

C. $y=\tfrac23$

D. $y=\tfrac32$

E. $y=-1$

Hint

For large $\lvert x\rvert$, compare the leading terms $-3x$ over $x$.

Solution

As $x\to\pm\infty$, $\frac{2-3x}{x+1}\to\frac{-3x}{x}=-3$. **Answer A.**

23 Range of $\displaystyle p(x)= \left( \frac{1}{3} \right)^x +2$

A. $[2,\infty)$

B. $(-\infty,2]$

C. $(-\infty,2)$

D. $(-\infty,\infty)$

E. $(2,\infty)$

Hint

The function $(\tfrac13)^x$ is always positive. What does that tell you about the range of $p(x)$? What happens if $x$ is very large or very small, negative?

Solution

Since $(\tfrac13)^x>0$ for all real $x$, \[ (\tfrac13)^x\in(0,\infty) \quad\Longrightarrow\quad p(x)=(\tfrac13)^x+2\in(2,\infty). \] \[ \boxed{(2,\infty)}. \] **Answer E.**

24 (***) Find $C$ and $b$ for $y = C\,b^x$

$x$	1	2	3	4
$y$	8	12	18	27

A. $C=3,\;b=2$

B. $C=\tfrac{16}{3},\;b=\tfrac23$

C. $C=8,\;b=\tfrac23$

D. $C=\tfrac{16}{3},\;b=\tfrac32$

E. $C=8,\;b=\tfrac32$

Hint

Evaluate the function at points we know.

Solution

First evaluate the function at $x=1$: \[ y(1)=C b^1\] This gives us the equation \[ 8 = C b.\] We can solve for $C$: \[ C = \frac{8}{b}.\] Next evaluate the function at $x=2$: \[ y(2)=C b^2\] This gives us the equation \[ 12 = C b^2.\]

Since we know $C$, we can substitute:

\[ 12 = \left(\frac{8}{b}\right) b^2 \]

Cancel the $b$ in the denominator: So we have

\[ 12 = 8b\]

Hence

\[ b = \frac{12}{8} = \frac{3}{2}.\]

Now substitute $b$ back into the equation for $C$:

\[ C = \frac{8}{b} = \frac{8}{\frac{3}{2}} = \frac{16}{3}.\]

So the answer is D.

25 Continuous Compounding

Arturo wants $15 000 in 6 years with continuous compounding at 3.7 % Continuously compounded. How much must he invest now?

A. $12 061.98

B. $1 629.14

C. $138 109.96

D. $18 728.57

E. $12 013.73

Hint

Use $A=Pe^{rt}$ with the given data.

Solution

We want to solve for $P$ in the equation \[ A=Pe^{rt}\] with data given $A=15000$, $r=\frac{3.7}{100}=0.037$, and $t=6$: \[ 15000=Pe^{0.037\cdot6}\] \[ P=\frac{15000}{e^{6(0.037)}}\approx\frac{15000}{e^{0.222}}=\frac{15000}{1.248}\approx12013.73.\] So the answer is $P\approx12013.73$. **Answer E.**

26 Population Prediction

A city’s population is $P(t)=12500e^{0.02t}$, where $t$ is years after 1985. Predict in 2020.

A. < 10 000

B. 10 000–20 000

C. 20 000–30 000

D. 30 000–40 000

E. > 40 000

Hint

Set $t=2020-1985=35$; compute $P(35)$.

Solution

In 1985, $t=0$. In 2020, $t=35$. Substituting $t=35$ into the population formula: \[ P(35)=12500\,e^{0.7}\approx12500\times2.0138\approx25\,171, \]

which lies between 20 000 and 30 000.
Answer C.

27 Evaluate $\log_{2}\!\bigl(\tfrac18\bigr)$

A. $\tfrac13$ B. $-3$ C. $\tfrac14$ D. $-4$ E. 0

Hint

Rewrite $\tfrac18$ as a power of $2$.

Solution

$\tfrac18=2^{-3}$, so \[ \log_{2}\!\bigl(\tfrac18\bigr)=\log_{2}\!\bigl(2^{-3}\bigr)=-3.\] Thus the answer is $-3$. **Answer B.**

28 Rewrite $6^{N}=13$ in Logarithmic Form

A. $\log_{6}13 = N$

B. $\log 6 = 13^{N}$

C. $\log_{13}6 = N$

D. $\log_{13}6 = N$

E. $\log_{6}13 = N$

Hint

Look at the answers and see which base logarithm makes sense to try.

Solution

Remember the definition of logarithm with base $b$: \[ \log_{b}(x)\] asks to what number do we raise $b$ to get $x$? so to put this in logarithm form it makes sense to use base 6. and recast the equation $6^{N}=13$ as \[ \log_{6} 6^{N} = \log_{6} 13\] The logarithm is the inverse of the exponentiation, so we can cancel the $6$ on the left side to obtain. \[ N = \log_{6} 13.\] Thus the answer is **E.**

29 Solve $\log_{2}(3)=\log_{2}(5x)-\log_{2}(x+1)$

A. $x=\tfrac12$ B. $x=\tfrac14$ C. $x=\tfrac32$ D. $x=\tfrac38$

Hint

Review the rules for combining logarithms.

Solution

Combine logs: \[ \log_{2}(3)=\log_{2}\!\left(\frac{5x}{x+1}\right).\] Exponentiate both sides: \[ 3 = \frac{5x}{x+1}.\] Multiply both sides by $x+1$: \[ 3(x+1) = 5x.\] Distribute: \[ 3x + 3 = 5x.\] Subtract $3x$ from both sides: \[ 3 = 5x - 3x\] \[ 3 = 2x.\] Divide both sides by $2$: \[ x = \frac{3}{2}.\] Thus the answer is $x=\frac{3}{2}$. **Answer C.**

30 Solve for $t$

$$ 140\Bigl(\tfrac12\Bigr)^{t/4} = 350 $$

A. $t = 4\log_{1/2}(5)$

B. $t = 4\log_{1/2}\!(\tfrac52)$

C. $t = 4\log_{5/2}(\tfrac12)$

D. $t = 4\log_{5}(2)$

Solution

We want to solve

$$ 140\Bigl(\tfrac12\Bigr)^{t/4} = 350. $$

Divide both sides by 140:

$$ \Bigl(\tfrac12\Bigr)^{t/4} = \frac{350}{140} = \frac{5}{2}. $$

Now take logarithms in base $\tfrac12$:

$$ \frac{t}{4} = \log_{1/2}\!\Bigl(\tfrac{5}{2}\Bigr) \quad\Longrightarrow\quad t = 4\,\log_{1/2}\!\Bigl(\tfrac{5}{2}\Bigr). $$

That matches choice B.

Answer B.

31 Solve $ \displaystyle \frac{8}{1 + 100e^{-3t}} = 2$

A. $\displaystyle t = -\ln\!\bigl(\tfrac{1}{100}\bigr)$

B. $\displaystyle t = \tfrac{1}{3}\ln\!\bigl(\tfrac{1}{20}\bigr)$

C. $\displaystyle t = -\ln\!\bigl(\tfrac{3}{20}\bigr)$

D. $\displaystyle t = -3\ln\!\bigl(\tfrac{3}{100}\bigr)$

E. $\displaystyle t = -\tfrac{1}{3}\ln\!\bigl(\tfrac{3}{100}\bigr)$

Solution

We want $t$ such that

$$ \frac{8}{1 + 100e^{-3t}} = 2. $$

Step by step:

$$ \begin{aligned} \frac{8}{1 + 100e^{-3t}} &= 2 &&\bigl|\times(1 + 100e^{-3t})\\ 8 &= 2\,(1 + 100e^{-3t}) &&\bigl|\;\text{distribute}\bigr.\\ 8 &= 2 + 200e^{-3t} &&\bigl|\;-2\bigr.\\ 6 &= 200\,e^{-3t} &&\bigl|\;\div200\bigr.\\ \frac{6}{200} &= e^{-3t} &&\bigl|\;\ln\bigr.\\ \ln\!\bigl(\tfrac{6}{200}\bigr) &= -3t &&\bigl|\;\div(-3)\bigr.\\ t &= -\tfrac{1}{3}\,\ln\!\bigl(\tfrac{6}{200}\bigr). \end{aligned} $$

Since $\tfrac{6}{200} = \tfrac{3}{100}$,

$$ \boxed{t = -\frac{1}{3}\ln\!\bigl(\tfrac{3}{100}\bigr)}, $$

which corresponds to choice E.

32 Doubling Time with Monthly Compounding

Suppose Matt initially invests $3000 in an account bearing 4 % interest compounded monthly. How long will it take for the deposit to double in value, rounded to the nearest 0.01 years?

A. 17.33 years

B. 17.36 years

C. 19.22 years

D. 19.74 years

E. None of these

Hint

Use the formula for monthly compounding: \[ A = P\Bigl(1 + \frac{r}{n}\Bigr)^{nt}, \] with $A=6000$, $P=3000$, $r=0.04$, $n=12$. Solve for $t$.

Solution

\[ 6000 = 3000\Bigl(1+\tfrac{0.04}{12}\Bigr)^{12t} \;\Longrightarrow\; 2 = \Bigl(1.003333\Bigr)^{12t} \;\Longrightarrow\; 12t = \frac{\ln2}{\ln(1.003333)} \;\Longrightarrow\; t = \frac{1}{12}\,\frac{\ln2}{\ln(1.003333)} \approx 17.36. \] **Answer B.**

33 Logistic Growth of Fish Population

A certain lake is stocked with 1000 fish. The population grows according to the logistic curve

\[ P(t)=\frac{10\,000}{1 + 9e^{-t/5}}, \]

where $t$ is months since stocking. After how many months will the fish population be 2000?

A. Less than 2 months

B. Between 2 and 4 months

C. Between 4 and 6 months

D. Between 6 and 8 months

E. More than 8 months

Hint

Set $P(t)=2000$ and solve for $t$: \[ 2000=\frac{10000}{1+9e^{-t/5}}. \]

Solution

\[ 2000(1+9e^{-t/5})=10000 \;\Longrightarrow\; 1+9e^{-t/5}=5 \;\Longrightarrow\; 9e^{-t/5}=4 \;\Longrightarrow\; e^{-t/5}=\frac{4}{9} \;\Longrightarrow\; -t/5=\ln\!\tfrac{4}{9} \;\Longrightarrow\; t = -5\ln\!\tfrac{4}{9}\approx 4.05. \] This is between 4 and 6 months. **Answer C.**

34 (***) Atmospheric‑Pressure Model

Atmospheric pressure $y$ decays exponentially with height $t$ (in feet) via

\[ y = C e^{kt}. \]

If at $t=18000$ ft the pressure is half of sea‑level ($y=\tfrac12C$), find $k$ and then estimate the pressure at $t=1000$ ft as a percentage of sea‑level.

A. Less than 20 %

B. 20–40 %

C. 40–60 %

D. 60–80 %

E. More than 80 %

Hint

1. Use $C e^{k\cdot18000}=\tfrac12C$ to find $k$. 2. Compute $e^{k\cdot1000}\times100\%$.

Solution

1. From $Ce^{18000k}=\tfrac12C$: \[ e^{18000k}=\tfrac12 \;\Longrightarrow\; 18000k=\ln\tfrac12=-\ln2 \;\Longrightarrow\; k=-\tfrac{\ln2}{18000}\approx -3.85\times10^{-5}. \] 2. At $t=1000$: \[ \frac{y}{C}=e^{k\cdot1000} =e^{-3.85\times10^{-5}\times1000} =e^{-0.0385}\approx0.9622, \] i.e.\ about 96.22 %. That is **more than 80 %**. **Answer E.**

35 Smartphone Users Model

The number of smartphone users (in millions) in the U.S. from 2010 to 2016 is

\[ f(t)=75\ln(t+1)+50, \]

where $t$ is years since 2010. According to this model, in what year did the number of users reach 150 million?

A. 2010

B. 2011

C. 2012

D. 2013

E. 2014

Hint

Set $75\ln(t+1)+50=150$ and solve for $t$, then add to 2010.

Solution

\[ 75\ln(t+1)+50=150 \;\Longrightarrow\; 75\ln(t+1)=100 \;\Longrightarrow\; \ln(t+1)=\tfrac{100}{75}=\tfrac{4}{3} \;\Longrightarrow\; t+1=e^{4/3} \;\Longrightarrow\; t=e^{4/3}-1\approx2.79. \] Year $=2010+2.79\approx2012.79$, i.e.\ late 2012. **Answer C.**

Answers

Problem	Answer
1	A
2	A
3	B
4	A
5	E
6	A
7	B
8	C
9	D
10	E
11	A
12	B
13	C
14	E
15	D
16	C
17	C
18	B
19	A
20	D
21	E
22	A
23	E
24	D
25	E
26	C
27	B
28	E
29	C
30	B
31	E
32	B
33	C
34	E
35	C

Exponential Growth / Decay	Logs Conversion
\(A = P\!\bigl(1+\tfrac{r}{n}\bigr)^{nt}\)	\(\log_b x = \dfrac{\ln x}{\ln b}\)
\(A = P e^{rt}\)

Problem	Answer
1	A
2	A
3	B
4	A
5	E
6	A
7	B
8	C
9	D
10	E
11	A
12	B
13	C
14	E
15	D
16	C
17	C
18	B
19	A
20	D
21	E
22	A
23	E
24	D
25	E
26	C
27	B
28	E
29	C
30	B
31	E
32	B
33	C
34	E
35	C

Form
\(y = a(x-p)(x-q)\)
\(y = a(x-h)^2 + k\)
\(y = a(x-r_1)(x-r_2)\)

Problem	Answer
1	A
2	A
3	B
4	A
5	E
6	A
7	B
8	C
9	D
10	E
11	A
12	B
13	C
14	E
15	D
16	C
17	C
18	B
19	A
20	D
21	E
22	A
23	E
24	D
25	E
26	C
27	B
28	E
29	C
30	B
31	E
32	B
33	C
34	E
35	C

MATH 112 – Final Exam Practice

2025/05/03

Formulas Provided on the Exam {: #formula-sheet }

Quadratics

Exponentials & Logs

Compound Interest

Problems

1 Sales‑Tax Expression

2 Which Graph is a Function?

3 Domain of \(R(x)=\sqrt{2-5x}\)

4 Intervals of Increase

5 Starbucks — Slope & Interpretation

6 Linear Model for Starbucks Locations

7 Evaluate the Piecewise Function

8 (*) Transformations of \(g(x)\)

9 Fuel‑Efficiency Scaling

10. (***) Composition of Rational Functions

11 (*) Coupon Order Matters

12 (*) Inverse Function

13 (*) Evaluate \(f^{-1}(2)\) from a Graph

14 Find a Quadratic from a Graph

15 Time for Projectile to Land

16 (*) Profit Maximization

17 Facebook Users Model

18 (***) When Did Users Reach 1 Billion?

19 Factor \(R(x)\) Given a Root

20 End Behavior of \( f(t)=\dfrac{6t}{t + 1.2} \)

21 (***) Evaluate \(h^{-1}(5)\) and Interpret

22 (***) Horizontal Asymptote of \(\displaystyle p(x)=\frac{2-3x}{x+1}\)

23 Range of \(\displaystyle p(x)= \left( \frac{1}{3} \right)^x +2\)

24 (***) Find \(C\) and \(b\) for \(y = C\,b^x\)

25 Continuous Compounding

26 Population Prediction

27 Evaluate \(\log_{2}\!\bigl(\tfrac18\bigr)\)

28 Rewrite \(6^{N}=13\) in Logarithmic Form

29 Solve \(\log_{2}(3)=\log_{2}(5x)-\log_{2}(x+1)\)

30 Solve for \(t\)

31 Solve \( \displaystyle \frac{8}{1 + 100e^{-3t}} = 2\)

32 Doubling Time with Monthly Compounding

33 Logistic Growth of Fish Population

34 (***) Atmospheric‑Pressure Model

35 Smartphone Users Model

Answers

1 Sales‑Tax Expression

2 Which Graph is a Function?

3 Domain of \(R(x)=\sqrt{2-5x}\)

4 Intervals of Increase

5 Starbucks — Slope & Interpretation

6 Linear Model for Starbucks Locations

7 Evaluate the Piecewise Function

8 (*) Transformations of \(g(x)\)

9 Fuel‑Efficiency Scaling

11 (*) Coupon Order Matters

12 (*) Inverse Function

13 (*) Evaluate \(f^{-1}(2)\) from a Graph

14 Find a Quadratic from a Graph

15 Time for Projectile to Land

16 (*) Profit Maximization

17 Facebook Users Model

18 (***) When Did Users Reach 1 Billion?

19 Factor \(R(x)\) Given a Root

20 End Behavior of \( f(t)=\dfrac{6t}{t + 1.2} \)

21 (***) Evaluate \(h^{-1}(5)\) and Interpret

22 (***) Horizontal Asymptote of \(\displaystyle p(x)=\frac{2-3x}{x+1}\)

23 Range of \(\displaystyle p(x)= \left( \frac{1}{3} \right)^x +2\)

24 (***) Find \(C\) and \(b\) for \(y = C\,b^x\)

25 Continuous Compounding

26 Population Prediction

27 Evaluate \(\log_{2}\!\bigl(\tfrac18\bigr)\)

28 Rewrite \(6^{N}=13\) in Logarithmic Form

29 Solve \(\log_{2}(3)=\log_{2}(5x)-\log_{2}(x+1)\)

30 Solve for \(t\)

31 Solve \( \displaystyle \frac{8}{1 + 100e^{-3t}} = 2\)

32 Doubling Time with Monthly Compounding

33 Logistic Growth of Fish Population

34 (***) Atmospheric‑Pressure Model

35 Smartphone Users Model

Problem	Answer
1	A
2	A
3	B
4	A
5	E
6	A
7	B
8	C
9	D
10	E
11	A
12	B
13	C
14	E
15	D
16	C
17	C
18	B
19	A
20	D
21	E
22	A
23	E
24	D
25	E
26	C
27	B
28	E
29	C
30	B
31	E
32	B
33	C
34	E
35	C