PAGE 1

MA 161 Exam 1

Tuesday, 2/1/2022, 6:30 pm

Recitation Instructor	Location
Isaac Chiu	CL50 224
Rowan Desjardins	CL50 224
Christopher Dewey	CL50 224
Hanan Gadi	WALC 1055
Mohit Pathak	WALC 1055
Vittal Srinivasan	PHYS 112
Ezekiel (Seun) Yinka Kehinde	PHYS 112

PAGE 2

The domain of the function \( \frac{\sqrt[4]{6 + 5x}}{x^2 - 9} \) is

A. \( x \neq \pm 3 \)
B. \( [-\frac{6}{5}, 3) \cup (3, \infty) \)
C. \( (-\infty, \frac{6}{5}] \)
D. \( (-\infty, -3) \cup (-3, \frac{6}{5}) \)
E. \( (3, \infty) \)

Solution Steps

no division by zero
no negative under even root

\( x^2 - 9 \neq 0 \rightarrow x \neq -3, x \neq 3 \)

\( 6 + 5x \geq 0 \)

\( 5x \geq -6 \rightarrow x \geq -6/5 \)

Figure: Number line with points -3, -6/5, and 3. A solid line starts at -6/5 and goes right, with an X at 3.

\( [-6/5, 3) \cup (3, \infty) \)

PAGE 3

9) Function Transformations

The graph of \( f(x) = 3^{x-4} + 1 \) is obtained from the graph of \( g(x) = 3^x \) by:

A) Shifting horizontally 4 units left and vertically 1 unit down.
B) Reflecting it about the \( x \)-axis and shifting vertically 1 unit up.
C) Shifting vertically 4 units up and horizontally 1 unit left.
D) Shifting horizontally 4 units to the right and vertically 1 unit up.
E) Shifting vertically 4 units down and horizontally 1 unit right.

Step-by-Step Analysis

\[ f(x) = 3^{x-4} + 1 \]

Horizontal Transformation

Change \( x \): horizontal shift

\( -4 \rightarrow \) RIGHT 4 units

Vertical Transformation

Change to \( y \): vertical shift

\( +1 \rightarrow \) UP 1 unit

PAGE 4

Finding the Inverse Function

The inverse of the function \( f(x) = \frac{3x - 2}{2x + 5} \) is \( f^{-1}(x) = \)

A. \( \frac{5x - 2}{3 - 2x} \)
B. \( \frac{2x - 5}{3 - 2x} \)
C. \( \frac{2x + 3}{5 - 2x} \)
D. \( \frac{5x + 2}{3 - 2x} \)
E. \( \frac{3x - 2}{3 - 5x} \)

Solution Process

\[ y = \frac{3x - 2}{2x + 5} \]

Interchange \( x \) and \( y \)

\[ x = \frac{3y - 2}{2y + 5} \]

Solve for \( y \)

\[ (x)(2y + 5) = 3y - 2 \]

\[ 2xy + 5x = 3y - 2 \]

\[ 2xy - 3y = -2 - 5x \]

\[ y(2x - 3) = -2 - 5x \]

\[ y = \frac{-2 - 5x}{2x - 3} = \frac{-(5x + 2)}{2x - 3} = \frac{5x + 2}{-(2x - 3)} = \frac{5x + 2}{3 - 2x} \]

PAGE 5

Trigonometric Composition Problem

For \(-1 \leq x \leq 1\), \(\tan(\sin^{-1} x)\) equals:

A. \(\sqrt{1 - x^2}\)
B. \(\frac{\sqrt{1 - x^2}}{x}\)
C. \(\frac{x}{\sqrt{1 - x^2}}\) (Correct)
D. \(x\sqrt{1 - x^2}\)
E. \(\frac{1}{\cos x}\)

Step-by-Step Solution

Let \(\theta = \sin^{-1} x\). This implies:

\(x = \sin \theta = \frac{\text{opp}}{\text{hyp}} = \frac{x}{1}\)

To find the tangent, we draw a triangle to determine the adjacent side.

Figure: Right triangle with hypotenuse 1, opposite side x, and adjacent side a = sqrt(1-x^2).

Solve for \(a\) using the Pythagorean Theorem:

\[1^2 = a^2 + x^2\]\[a^2 = 1 - x^2\]\[a = \sqrt{1 - x^2}\]

Now, calculate the tangent of the angle:

\[\tan(\theta) = \frac{\text{opp}}{\text{adj}} = \frac{x}{\sqrt{1 - x^2}}\]

PAGE 6

Difference Quotient Evaluation

If \(f(x) = \left(2 + \frac{3}{x}\right)\), then the expression \(\frac{f(a+h) - f(a-h)}{2h} = \)

\(f(a+h) = 2 + \frac{3}{a+h}\)

\(f(a-h) = 2 + \frac{3}{a-h}\)

A. \(1 + \frac{3}{a^2 - h^2}\)
B. \(2 + \frac{6}{a^2 - h^2}\)
C. \(\frac{1}{h} - \frac{1}{2(a^2 - h^2)}\)
D. \(-\frac{3}{a^2 - h^2}\) (Correct)
E. \(-\frac{6}{a^2 - h^2}\)

Algebraic Simplification

First, find the numerator:

\[f(a+h) - f(a-h) = 2 + \frac{3}{a+h} - \left(2 + \frac{3}{a-h}\right)\]\[= 2 + \frac{3}{a+h} - 2 - \frac{3}{a-h}\]\[= \frac{3}{a+h} - \frac{3}{a-h}\]

Find a common denominator:

\[= \frac{3}{a+h} \cdot \frac{a-h}{a-h} - \frac{3}{a-h} \cdot \frac{a+h}{a+h}\]\[= \frac{3a - 3h}{a^2 - h^2} - \frac{3a + 3h}{a^2 - h^2}\]\[= \frac{3a - 3h - (3a + 3h)}{a^2 - h^2} = \frac{3a - 3h - 3a - 3h}{a^2 - h^2}\]\[= \frac{-6h}{a^2 - h^2}\]

Now, divide by \(2h\):

\[\frac{f(a+h) - f(a-h)}{2h} = \frac{-6h}{a^2 - h^2} \cdot \frac{1}{2h} = \frac{-3}{a^2 - h^2}\]

PAGE 7

Limit Evaluation: Factoring and Cancellation

\[ \lim_{x \to 2} \frac{2x - x^2}{x^2 - x - 2} = \]

Multiple Choice Options

A. \(-\frac{2}{3}\) (Correct)
B. \(\frac{2}{3}\)
C. 0
D. \(\infty\)
E. \(-\infty\)

Step 1: Direct Substitution

Try \(x = 2\):

\[ \frac{2(2) - (2)^2}{(2)^2 - (2) - 2} = \frac{0}{0} \]

Strategy: Factor & Cancel

Since we have an indeterminate form, we must factor the numerator and denominator.

Step 2: Factoring and Simplifying

\[ \lim_{x \to 2} \frac{x(2 - x)}{(x - 2)(x + 1)} \]

Note: Reverse order: change sign

\[ = \lim_{x \to 2} \frac{-x(x - 2)}{(x - 2)(x + 1)} \]

Canceling the common factor \((x - 2)\):

\[ = \lim_{x \to 2} \frac{-x}{x + 1} \]

Step 3: Final Evaluation

Try \(x = 2\) again:

\[ = \frac{-2}{2 + 1} = -\frac{2}{3} \]

PAGE 8

Limit Evaluation: Rationalization

\[ \lim_{x \to 3} \frac{x - 3}{\sqrt{x} - \sqrt{3}} \]

Multiple Choice Options

a. \(2\sqrt{3}\) (Correct)
b. \(\frac{\sqrt{3}}{3}\)
c. \(\frac{2\sqrt{3}}{3}\)
d. \(\sqrt{3}\)
e. does not exist

Step 1: Direct Substitution

Try \(x = 3\):

\[ \frac{3 - 3}{\sqrt{3} - \sqrt{3}} = \frac{0}{0} \]

Strategy: Rationalize

Roots are involved: multiply by the conjugate.

Step 2: Rationalizing the Denominator

\[ \lim_{x \to 3} \frac{x - 3}{\sqrt{x} - \sqrt{3}} \cdot \frac{\sqrt{x} + \sqrt{3}}{\sqrt{x} + \sqrt{3}} \]

\[ = \lim_{x \to 3} \frac{(x - 3)(\sqrt{x} + \sqrt{3})}{(\sqrt{x} - \sqrt{3})(\sqrt{x} + \sqrt{3})} = \lim_{x \to 3} \frac{(x - 3)(\sqrt{x} + \sqrt{3})}{(x - 3)} \]

Canceling the common factor \((x - 3)\):

\[ = \lim_{x \to 3} \sqrt{x} + \sqrt{3} \]

Step 3: Final Evaluation

Try \(x = 3\) again:

\[ = \sqrt{3} + \sqrt{3} = 2\sqrt{3} \]

PAGE 9

\[ \lim_{x \to 2^-} \frac{x+1}{x-2} = \]

try \( x = 2 \) :

\[ \frac{3}{0} \to \infty \text{ or } -\infty \]

depending on the sign of the ratio

means a small number can be positive or negative

A. \( \infty \)
B. \( -\infty \) (Correct)
C. 0
D. 3
E. 1

\( x \to 2^- \) means \( x \) is a number close to 2 but less than 2

let \( x = 1.9999 \)

then

\[ \frac{x+1}{x-2} = \frac{2.9999}{-0.0001} \]

overall ratio is negative

\[ \text{so } \lim_{x \to 2^-} \frac{x+1}{x-2} = -\infty \]

PAGE 10

Spring 2020

Problem 7

7. Find all \( x \) in \( [0, 2\pi] \) such that \( 2 \sin x \cos x + \cos x = 0 \)

\[ 2 \sin x \cos x = -\cos x \]

if you divide by \( \cos x \)

\[ 2 \sin x = -1 \]

this will only give part of the solutions

means \( \cos x \neq 0 \)

but \( \cos x \) could be zero on \( [0, 2\pi] \)

right way:

\[ 2 \sin x \cos x + \cos x = 0 \]\[ (\cos x)(2 \sin x + 1) = 0 \]

\( \cos x = 0 \)

\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \]

or \( 2 \sin x + 1 = 0 \)

\( \sin x = -1/2 \)

\[ x = \frac{7\pi}{6}, \frac{11\pi}{6} \]

PAGE 11

Fall 2021 #8 ignore (derivative question)

Spring 2019 #9

\[ \lim_{x \to 2} \frac{\cos\left(\frac{x^2-4}{\pi}\right)(x-2)}{\sqrt{x^2+12}-4} \]

Note: \(\cos\left(\frac{x^2-4}{\pi}\right)\) goes to 1 as \(x \to 2\).

By plugging in \(x=2\):

\[ \frac{\cos(0)(0)}{\sqrt{16}-4} \to \frac{0}{0} = ? \]

The \(\cos\left(\frac{x^2-4}{\pi}\right)\) goes to 1, does not affect limit.

Focus on:

\[ \lim_{x \to 2} \frac{x-2}{\sqrt{x^2+12}-4} \cdot \frac{\sqrt{x^2+12}+4}{\sqrt{x^2+12}+4} \]

\[ = \lim_{x \to 2} \frac{(x-2)(\sqrt{x^2+12}+4)}{(\sqrt{x^2+12})^2 - (4)^2} = \lim_{x \to 2} \frac{(x-2)(\sqrt{x^2+12}+4)}{x^2+12-16} \]

\[ = \lim_{x \to 2} \frac{(x-2)(\dots)}{x^2-4} = \lim_{x \to 2} \frac{(x-2)(\sqrt{x^2+12}+4)}{(x+2)(x-2)} \]

\[ = \lim_{x \to 2} \frac{\sqrt{x^2+12}+4}{x+2} = \frac{\sqrt{16}+4}{4} = \frac{8}{4} = 2 \]

Exam Locations and Instructors

Solution Steps

9) Function Transformations

Step-by-Step Analysis

Horizontal Transformation

Vertical Transformation

Finding the Inverse Function

Solution Process

Trigonometric Composition Problem

Step-by-Step Solution

Solve for \(a\) using the Pythagorean Theorem:

Difference Quotient Evaluation

Algebraic Simplification

Limit Evaluation: Factoring and Cancellation

Multiple Choice Options

Step 1: Direct Substitution

Step 2: Factoring and Simplifying

Step 3: Final Evaluation

Limit Evaluation: Rationalization

Multiple Choice Options

Step 1: Direct Substitution

Step 2: Rationalizing the Denominator

Step 3: Final Evaluation

right way:

Focus on: