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MA 161 Exam 1

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

Exam Locations and Instructors

Recitation InstructorLocation
Isaac ChiuCL50 224
Rowan DesjardinsCL50 224
Christopher DeweyCL50 224
Hanan GadiWALC 1055
Mohit PathakWALC 1055
Vittal SrinivasanPHYS 112
Ezekiel (Seun) Yinka KehindePHYS 112
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The domain of the function \( \frac{\sqrt[4]{6 + 5x}}{x^2 - 9} \) is

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

Solution Steps

  • no division by zero
  • no negative under even root

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

\( 6 + 5x \geq 0 \)

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

Number line with points -3, -6/5, and 3. A solid line starts at -6/5 and goes right, with an X at 3.
\( [-6/5, 3) \cup (3, \infty) \)
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9) Function Transformations

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

  1. A) Shifting horizontally 4 units left and vertically 1 unit down.
  2. B) Reflecting it about the \( x \)-axis and shifting vertically 1 unit up.
  3. C) Shifting vertically 4 units up and horizontally 1 unit left.
  4. D) Shifting horizontally 4 units to the right and vertically 1 unit up.
  5. 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

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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 \( y \)

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

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

  1. A. \(\sqrt{1 - x^2}\)
  2. B. \(\frac{\sqrt{1 - x^2}}{x}\)
  3. C. \(\frac{x}{\sqrt{1 - x^2}}\) (Correct)
  4. D. \(x\sqrt{1 - x^2}\)
  5. 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.

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:

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

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

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

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

  1. A. \(1 + \frac{3}{a^2 - h^2}\)
  2. B. \(2 + \frac{6}{a^2 - h^2}\)
  3. C. \(\frac{1}{h} - \frac{1}{2(a^2 - h^2)}\)
  4. D. \(-\frac{3}{a^2 - h^2}\) (Correct)
  5. 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 \(2h\):

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

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

\[ = \sqrt{3} + \sqrt{3} = 2\sqrt{3} \]
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\[ \lim_{x \to 2^-} \frac{x+1}{x-2} = \]

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

  1. A. \( \infty \)
  2. B. \( -\infty \) (Correct)
  3. C. 0
  4. D. 3
  5. 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

\[ \text{so } \lim_{x \to 2^-} \frac{x+1}{x-2} = -\infty \]
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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 \( 2 \sin x + 1 = 0 \)

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

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

\[ \lim_{x \to 2} \frac{x-2}{\sqrt{x^2+12}-4} \cdot \frac{\sqrt{x^2+12}+4}{\sqrt{x^2+12}+4} \]
\[ = \lim_{x \to 2} \frac{(x-2)(\sqrt{x^2+12}+4)}{(\sqrt{x^2+12})^2 - (4)^2} = \lim_{x \to 2} \frac{(x-2)(\sqrt{x^2+12}+4)}{x^2+12-16} \]
\[ = \lim_{x \to 2} \frac{(x-2)(\dots)}{x^2-4} = \lim_{x \to 2} \frac{(x-2)(\sqrt{x^2+12}+4)}{(x+2)(x-2)} \]
\[ = \lim_{x \to 2} \frac{\sqrt{x^2+12}+4}{x+2} = \frac{\sqrt{16}+4}{4} = \frac{8}{4} = 2 \]