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Exam Review

$\gdef \N{\mathbb{N}} \gdef \Z{\mathbb{Z}} \gdef \Q{\mathbb{Q}} \gdef \R{\mathbb{R}} \gdef \C{\mathbb{C}} \gdef \setcomp#1{\overline{#1}} \gdef \sseq{\subseteq} \gdef \pset#1{\mathcal{P}(#1)} \gdef \covariant{\operatorname{Cov}} \gdef \of{\circ} \gdef \p{^{\prime}} \gdef \pp{^{\prime\prime}} \gdef \ppp{^{\prime\prime\prime}} \gdef \pn#1{^{\prime\times{#1}}} $

Below are the types of questions asked on exams. This is a near-complete list of questions that have been asked on exams in the past. If you can answer all of these questions, you should be able to answer any question on the exam. However, this is not a complete list of all possible questions, so don’t rely on this alone. If you encounter a question that is not on this list, please add it using the edit button near the top of the page.

Best of luck, I’m rooting for ya!

Exam Questions

  • What is the domain of…
  • Find all asymptotes of…
  • Compute the integral of…
  • Where is \(f\) differentiable, cts, etc
  • What value for a piecewise function makes it cts
  • Verbatim memorization of theorems (I know, crazy.)
  • Proofs of theorems and corollaries
  • Find the bounded area
  • Limit calculation
  • Finding limits by definition of limit
  • Finding tangent line to a point on a function
  • Find regions of ascent and descent
  • Name the type of discontinuity at a point
  • Sketch the function
  • Computing single sided limits
  • Find where a function is convex/concave
  • Find local extreme points
\[\gdef \bm{\boldsymbol}\]
Finding the domain of a function

The domain of a function is the set of values for which the function is defined.
At any point where a component of the function is undefined, the entire function is undefined as well.

  • Polynomials are defined everywhere
  • Rationals are defined where the denominator is not zero
  • Exponents are defined everywhere except where the exponent in simplified form has an even denominator
  • The Logarithm function is defined for \(x>0\)
Identifying Asymptotes of a Function
Shortcut for Simple Rational Functions
Horizontal Asymptotes
  • If the numerator and denominator have an equal degree
    • The equation for the asymptote is \(\frac{\text{leading coefficient of the numerator}}{\text{leading coefficient of the denominator}}\)
  • If the degree of the numerator is less than the denominator
    • The asymptote is \(y=0\)
  • If the degree of the numerator is greater than the denominator
    • There is no horizontal asymptote
Diagonal (Slant) Asymptotes

A rational function has a slant asymptote only if the degree of the numerator is exactly one more than that of the denominator. The equation is obtained by dividing the numerator by the denominator and discarding the remainder.

Vertical Asymptotes

Vertical asymptotes occur where factors in the denominator that are not canceled out by the numerator equal zero. You can identify these asymptotes by factoring the numerator and denominator. Any matching terms in the top and bottom are removable discontinuities. Any unmatched term in the denominator will cause a vertical asymptote when that term is equal to zero. The equation for that asymptote is: \(x=\text{the value that makes the term zero}\).

Identifying Asymptotes of Any Function:

When dealing with a non rational function, you can identify the asymptotes by evaluating limits at key values:

Horizontal Asymptotes

If \(\lim_{x\to\infty}{f(x)}=L\) or \(\lim_{x\to-\infty}{f(x)}=L\), we say that \(L\) is a horizontal asymptote of \(f\).
Basically, evaluate the limit of the function as it approaches either + or - ∞. If it is a finite number, that’s your horizontal asymptote.

Diagonal Asymptotes

If you have a horizontal asymptote from the right, you don’t need to check as \(x\to\infty\).
If you have a horizontal asymptote from the left, you don’t need to check as \(x\to-\infty\).
If you have both, there is no diagonal asymptote.
To check if a diagonal asymptote exists, and if so, find it, for a given function \(f\):
WLOG \(\pm\infty\)

  1. Let \(a=\lim_{x\to\infty}{\frac{f(x)}x}\). If this does not equate to a finite number, there is no diagonal asymptote as \(x\to\infty\)
  2. Let \(b=\lim_{x\to\infty}{f(x)-ax}\). If this does not equate to a finite number, there is no diagonal asymptote as \(x\to\infty\).
  3. Check from the other side, if needed.

The equation for the asymptote is: \(y=ax+b\).

Vertical Asymptotes

If \(\lim_{x\to c}{f(x)}=\pm\infty\), we say the function has a vertical asymptote at \(c\).
A good method for identifying candidates for vertical asymptotes is to find values that are at the edge of the domain for a function. Also, any terms in any denominator that can go to zero for a particular value of \(x\) must be checked.

Computing the Integral

There’s not really a straightforward how-to for this, make sure you’re familiar with 05. Indefinite Integrals and 06. Definite Integrals.

Finding Where \(f\) Is Continuous

\(f(x)\text{ is continuous at }x=a\text{ if }f(a)=\lim_{x\to a}{f(x)}\).

3 Types of Discontinuities:
  1. \(\lim_{x\to a}{f(x)}\) exists, but \(\not=f(a)\). Either \(f(a)\) is undefined, or some other number.
    • This is called a removable discontinuity, because it’s an issue at just that point.
  2. \(\lim_{x\to a}{f(x)}\) doesn’t exist because \(\lim_{x\to a^-}{f(x)} \not=\lim_{x\to a^+}{f(x)}\)
    • This is called a jump discontinuity, or a Type I discontinuity.
  3. \(\lim_{x\to a}{f(x)}\) doesn’t exist, even one-sided (from at least one side).
    • This is called a Type II discontinuity.
    • This does not include extended limits, so by this definition, \(\lim_{x\to0}{\frac1x}\) doesn’t have a limit, meaning it’s discontinuous at \(x=0\) with a Type II discontinuity.
    • Usually in this case, \(f\) has a vertical asymptote \(x=a\), but not always.
Finding Discontinuities:

See 02. Continuity for more details
- Rational Polynomials can be, but are not definitely, discontinuous where the denominator equals zero
- Piecewise functions can be discontinuous between the definitions
- Exponential functions can be discontinuous when the denominator of the exponent contains x.

Finding a value \(A\) to make \(f\) continuous at \(x_0\)

This is a relatively easy question to solve. You take the limit of \(f\) from both sides as \(x\to x_0\). If the two limits are finite and equal, that number is your \(A\). If they don’t equal or are not finite, there is no \(A\).

Finding where \(f\) is differentiable
  1. \(f(x)\) is differentiable at the point \(x=x_0\) if the limit \(\lim_{h\to 0}{\frac{f(x_0+h)-f(x_0)}{h}}\) exists.
  2. \(f(x)\) is differentiable from the left if \(\lim_{h\to 0^-}{\frac{f(x_0+h)-f(x_0)}{h}}\) exists.
  3. \(f(x)\) is differentiable from the right if \(\lim_{h\to 0^+}{\frac{f(h_0+h)-f(x_0)}{h}}\) exists.

    • \(f\) is differentiable at \(x_0\) \(\iff\) \(f\) is differentiable from left and from right at \(x_0\) and these limits are equal

    • Example: Graph \(y=|x|\). Is x differentiable at \(x=0\)?
      From the left: \(\lim_{h\to 0^-}{\frac{f(h)-f(0)}{h}}\) = \(\lim_{h\to 0}{\frac{-h}{h}}=-1\). Note, \(h<0\), so \(|h|=-h\)
      From the right: \(\lim_{h\to 0^+}{\frac{f(h)-f(0)}{h}}=1\)
      Hence, \(f(x)=|x|\) is differentiable from each side at \(x_0=0\), but is not differentiable there.
      4. \(f\) is differentiable in open interval \((a,b)\) if f is differentiable at each pt. in that interval

Limit Calculation by Definition

Quick refresher on symbols:
In the limit: \(\lim_{x\to a}{f(x)}=L\):

\(\boldsymbol{\delta}\) : Delta is the value that represents the small difference between \(x\) and \(a\) \({(|x-a|)}\)
\(\boldsymbol{\varepsilon}\) : Epsilon is the value that represents the small difference between \(f(x)\) and \(L\) \({(|f(x)-L|)}\)

The formal definition is:

\[\lim_{x\to a}{f(x)}=L\text{ if, for any }\varepsilon>0,\text{ there exists a }\delta>0\text{ so that } |f(x)-L|<\varepsilon\text{ when }0<|x-a|<\delta\]

If this is gibberish to you, check out this website for clarification.

Now, on to actually using it:
We want to get from \(0<|x-a|<\delta\) to \(|f(x)-L|<\varepsilon\).
We start with \(|f(x)-L|\). Put in the values for \(f(x)\) and \(L\). Simplify and rearrange whatever you can to get to the form \(|x-a|<\text{ some expression involving }\varepsilon\).

That expression involving \(\varepsilon\) is your \(\delta\), and by showing that such a delta exists, you complete the proof.

Example: Prove that \(\lim_{x\to4}{}3x+5=17\)

Let \(\varepsilon>0\). We seek \(\delta>0\) such that if \(|x-4|<\delta\), then \(|(3x+5)-17|<\varepsilon\).
We work “backwards:”
\(\(\begin{aligned}&|(3x+5)-17|<\varepsilon\\ \implies&|3x-12|<\varepsilon\\ \implies&3|x-4|<\varepsilon\\ \implies&|x-4|<\frac\varepsilon3 \end{aligned}\)\)

\[\delta=\frac\varepsilon3\]
\[Q.E.D.\]
Extended Limits

When there’s a \(\infty\) in the limit, we have to tweak the definition we are proving with.

  • \(\lim_{x\to\pm\infty}{f(x)}=L\)
    • If \(x\to+\infty\): \(\forall\varepsilon>0,\;\exists N\text{ such that if }x\bm>N,\;|f(x)-L|<\varepsilon\)
    • If \(x\to-\infty\): \(\forall\varepsilon>0,\;\exists N\text{ such that if }x\bm<N,\;|f(x)-L|\bm<\varepsilon\)
  • \(\lim_{x\to a}{f(x)}=\pm\infty\)
    • If \(\lim=+\infty\): \(\forall M,\;\exists\delta>0\text{ such that if }|x-a|<\delta,f(x)>M\)
    • If \(\lim=-\infty\): \(\forall M,\;\exists\delta>0\text{ such that if }|x-a|<\delta,f(x)<M\)
Verbatim Memorization of Theorems

You must know the Mean Value Theorem and it’s corollaries. It is often referred to as Lagrange’s Theorem. This theorem states that if \(f\) is continuous in \([a,b]\) and differentiable in \((a,b)\), there exists at least one point on \(f\) between \((a,b)\) where the slope of the tangent is exactly equal to the slope of the line between \(a\) and \(b\).