Coverage: Lessons 1–10
Exam Date and Time: Wednesday, September 23, 8:00–9:00 pm
Locations: LILY 1105, MA 175, WALC 1055. Refer to the posted seating chart for your location.
Review of Functions [Lesson 1]
Functions, Domain, and Range
A function \(f\) assigns each input \(x\) a unique output \(f(x)\). The domain is the set of admissible inputs; the range is the set of outputs.
Two families are common enough to know by sight: the domain of a polynomial is all of \((-\infty,\infty)\), and the domain of a rational function \(f(x)=p(x)/q(x)\) is every \(x\) with \(q(x)\neq0\).
Vertical Line Test
A curve is the graph of a function if and only if every vertical line crosses it at most once.
Composition and Symmetry
In general, \(f(g(x)) \neq g(f(x))\). The composition \((f\circ g)(x)\) is defined precisely when the range of \(g\) is contained in the domain of \(f\) — always check this before composing.
Even function: \(f(-x)=f(x)\) — graph is symmetric about the \(y\)-axis.
Odd function: \(f(-x)=-f(x)\) — graph is symmetric about the origin.
Transformations of \(y=f(x)\)
| Transformation | Effect |
|---|---|
| \(y=f(x)+k\) | Vertical shift \(k\) units: up if \(k>0\), down if \(k<0\) |
| \(y=f(x-k)\) | Horizontal shift \(k\) units: right if \(k>0\), left if \(k<0\) |
| \(y=kf(x)\) | Vertical stretch if \(|k|>1\), compression if \(0<|k|<1\) |
| \(y=f(kx)\) | Horizontal compression if \(|k|>1\), stretch if \(0<|k|<1\) |
| \(y=f(-x)\) | Reflect across the \(y\)-axis |
| \(y=-f(x)\) | Reflect across the \(x\)-axis |
Practice Problems
Problem 1. Given the graph of \(y=f(x)\), describe how to obtain the graph of \(y=f(2x-6)+4\).
▼ Show answerWrite \(2x-6=2(x-3)\): compress the graph horizontally by a factor of 2, shift right 3 units, then shift up 4 units
Problem 2. Suppose the domain of \(f\) is \([-4,6]\) and the range of \(f\) is \([-2,5]\). Find the domain and range of \(g(x)=3f(2x-4)-1\).
▼ Show answerDomain \([0,5]\): \(g\) is defined exactly when \(-4\le 2x-4\le 6\), i.e. \(0\le x\le5\). Range \([-7,14]\): the outputs of \(f\) run over \([-2,5]\), and \(3y-1\) sends \([-2,5]\) to \([3(-2)-1,\,3(5)-1]\)
Problem 3. Determine whether \(f(x) = x^4 - 3x^2\) is even, odd, or neither.
▼ Show answerEven, since \(f(-x)=f(x)\)
Problem 4. Describe, in words, the sequence of transformations that produces \(y=-2f(x-3)+1\) from \(y=f(x)\).
▼ Show answerShift right 3, vertically stretch by a factor of 2 and reflect across the \(x\)-axis, then shift up 1
Problem 5. Find the domain of \(f(x) = \sqrt{4-x^2} + \dfrac{1}{x+1}\).
▼ Show answer\([-2,-1)\cup(-1,2]\)
Problem 6. Find the domain of \(f(x) = \dfrac{\ln(x^2-5x+6)}{\sqrt{9-x^2}}\).
▼ Show answer\((-3,2)\)
Problem 7. If \(f(x)=2x-1\) and \(f(g(x))=6x+5\), find \(g(x)\) and determine whether \(g\) is even, odd, or neither.
▼ Show answer\(g(x)=3x+3\); neither (\(g(-x)=-3x+3\) matches neither \(g(x)\) nor \(-g(x)\))
Inverse Functions and Logarithms [Lesson 2]
Horizontal Line Test
A function is one-to-one, and therefore invertible, if and only if every horizontal line crosses its graph at most once. If a function is not 1–1 on its whole domain, restrict the domain to a piece where it is.
Finding \(f^{-1}\)
By definition, \(y=f^{-1}(x) \iff f(y)=x\), which gives the round-trip identities \(f^{-1}(f(x))=x\) and \(f(f^{-1}(y))=y\). To find \(f^{-1}\) explicitly: (i) solve \(y=f(x)\) for \(x\) to get \(x=f^{-1}(y)\); (ii) interchange \(x\) and \(y\). The graph of \(f^{-1}\) is the reflection of the graph of \(f\) across \(y=x\).
Exponential and Logarithmic Functions
\(y=b^x\) (\(b>0,\ b\neq1\)): domain \((-\infty,\infty)\), range \((0,\infty)\)
\(y=\log_b x\): domain \((0,\infty)\), range \((-\infty,\infty)\). The natural log is \(\ln x=\log_e x\).
Law of Exponents: \(b^{x+y}=b^xb^y, \quad b^{x-y}=\dfrac{b^x}{b^y}, \quad (b^x)^r=b^{xr}\)
Log Rules
- \(\log_b(xy)=\log_b x+\log_b y\)
- \(\log_b(x/y)=\log_b x-\log_b y\)
- \(\log_b(x^r)=r\log_b x\)
- \(\log_b b^x=x, \quad b^{\log_b x}=x\)
- Change of base: \(\log_b x = \dfrac{\ln x}{\ln b}\)
Practice Problems
Problem 1. Find \(f^{-1}(x)\) if \(f(x) = \dfrac{2x+1}{3}\).
▼ Show answer\(f^{-1}(x) = \dfrac{3x-1}{2}\)
Problem 2. Solve \(3^{x-1} = 20\) for \(x\).
▼ Show answer\(x = 1+\dfrac{\ln 20}{\ln 3} \approx 3.727\)
Problem 3. Simplify \(\log_2(8x^3) - \log_2(2x)\) for \(x>0\).
▼ Show answer\(2+2\log_2 x\)
Problem 4. Show that \(f(x)=x^3+2\) is one-to-one and find \(f^{-1}(x)\).
▼ Show answer\(f\) is strictly increasing (its graph passes the horizontal line test), so it is one-to-one; \(f^{-1}(x) = (x-2)^{1/3}\)
Problem 5. Convert \(5^x\) to base \(e\).
▼ Show answer\(5^x = e^{x\ln 5}\)
Problem 6. Solve for \(x\): \(\log_2(x+3)+\log_2(x-1)=5\).
▼ Show answer\(x=5\) (the solution \(x=-7\) is rejected since it falls outside the domain)
Problem 7. Find \(f^{-1}(x)\) for \(f(x)=e^{2x-1}\), then evaluate \(f^{-1}(e^3)\).
▼ Show answer\(f^{-1}(x)=\dfrac{\ln x+1}{2}\); \(f^{-1}(e^3)=2\)
Trigonometry Review and Inverse Trig Functions [Lesson 3]
Right-Triangle Ratios and the Unit Circle
Unit circle: \((\cos\theta,\sin\theta)\) is the point at angle \(\theta\). Radians: \(2\pi\) rad \(=360^\circ\); arc length \(s=r\theta\).
Key Trig Identities
Double-angle / power-reduction: \(\cos2\theta=\cos^2\theta-\sin^2\theta\), \(\sin2\theta=2\sin\theta\cos\theta\), \(\cos^2\theta=\dfrac{1+\cos2\theta}{2}\), \(\sin^2\theta=\dfrac{1-\cos2\theta}{2}\)
General Sine Wave
For \(f(\theta)=a\sin(b(\theta-c))+d\): amplitude \(|a|\), period \(2\pi/|b|\), horizontal shift \(c\), vertical shift \(d\).
Inverse Trig Functions
| Function | Domain | Range |
|---|---|---|
| \(\sin^{-1}x\) | \([-1,1]\) | \([-\pi/2,\pi/2]\) |
| \(\cos^{-1}x\) | \([-1,1]\) | \([0,\pi]\) |
| \(\tan^{-1}x\) | \((-\infty,\infty)\) | \((-\pi/2,\pi/2)\) |
Reference-triangle technique: to simplify expressions like \(\sin(\cos^{-1}x)\), let \(\theta=\cos^{-1}x\), so \(\cos\theta=x/1\); draw a right triangle with adjacent \(x\) and hypotenuse \(1\), so the opposite side is \(\sqrt{1-x^2}\) by the Pythagorean theorem. Then \(\sin(\cos^{-1}x)=\sin\theta=\sqrt{1-x^2}\). The same idea — label the triangle from the inner inverse-trig definition, then read off the requested ratio — handles any composition like \(\cos(\tan^{-1}x)\).
Practice Problems
Problem 1. Find the exact value of \(\cos^{-1}\!\left(\cos\dfrac{6\pi}{5}\right)\).
▼ Show answer\(\dfrac{4\pi}{5}\) (since \(6\pi/5\) is outside \(\cos^{-1}\)'s range \([0,\pi]\), use \(\cos(6\pi/5)=\cos(4\pi/5)\), and \(4\pi/5\) is in range)
Problem 2. Find the exact value of \(\sin^{-1}\!\left(\sin\dfrac{6\pi}{5}\right)\).
▼ Show answer\(-\dfrac{\pi}{5}\) (since \(6\pi/5\) is outside \(\sin^{-1}\)'s range \([-\pi/2,\pi/2]\), use \(\sin(6\pi/5)=-\sin(\pi/5)=\sin(-\pi/5)\), and \(-\pi/5\) is in range)
Problem 3. Simplify \(\sin(\tan^{-1}x)\) for real \(x\).
▼ Show answer\(\dfrac{x}{\sqrt{1+x^2}}\)
Problem 4. Find the amplitude, period, and phase shift of \(f(\theta)=3\sin\!\left(2\theta-\dfrac{\pi}{3}\right)+1\).
▼ Show answerWrite \(2\theta-\pi/3=2(\theta-\pi/6)\): amplitude \(3\), period \(\pi\), phase (horizontal) shift \(\pi/6\) to the right, vertical shift \(+1\)
Problem 5. Evaluate \(\tan^{-1}(\sqrt3)\).
▼ Show answer\(\dfrac{\pi}{3}\)
Problem 6. Solve \(2\sin^2\theta - \sin\theta - 1 = 0\) for all \(\theta\) in \([0,2\pi)\).
▼ Show answer\(\theta = \dfrac{\pi}{2}, \dfrac{7\pi}{6}, \dfrac{11\pi}{6}\)
Problem 7. Simplify \(\cos(2\tan^{-1}x)\) to an algebraic expression in \(x\) (no trig functions).
▼ Show answer\(\dfrac{1-x^2}{1+x^2}\)
Velocity, Definition of Limit, One-Sided Limits [Lesson 4]
Average and Instantaneous Velocity
Instantaneous velocity at \(a\) is the limit of average velocity as the interval shrinks to \(a\).
The same quotient is also the slope of the secant line through \((a,f(a))\) and \((b,f(b))\): \(m_{\text{sec}}=\dfrac{f(b)-f(a)}{b-a}\). Letting \(b\to a\) turns the secant into the tangent line, with slope \(m_{\text{tan}}=\displaystyle\lim_{b\to a}\dfrac{f(b)-f(a)}{b-a}\) — the same limiting process as instantaneous velocity, just viewed geometrically.
Definition of a Limit
\(\lim_{x\to a}f(x)=L\) means \(f(x)\) gets arbitrarily close to \(L\) as \(x\) gets close to \(a\) (from either side), regardless of \(f(a)\) itself.
One-Sided Limits and Existence
The value \(f(a)\) is irrelevant to whether \(\lim_{x\to a}f(x)\) exists — only the behavior of \(f\) near \(a\) matters. If the one-sided limits disagree, the two-sided limit does not exist (DNE).
Practice Problems
Problem 1. For \(f(x)= \begin{cases} x+1, & x<2 \\ x^2-1, & x\ge2 \end{cases}\), does \(\lim_{x\to2}f(x)\) exist? If so, find it.
▼ Show answerYes, the limit is \(3\) (both one-sided limits equal 3)
Problem 2. Find the average velocity of \(s(t)=t^2+3t\) on \([1,4]\).
▼ Show answer\(8\)
Problem 3. Evaluate \(\displaystyle\lim_{x\to0}\dfrac{\sqrt{x+4}-2}{x}\).
▼ Show answer\(\dfrac14\)
Problem 4. Does \(\displaystyle\lim_{x\to3}f(x)\) exist for \(f(x)=\begin{cases}2x-1,&x\neq3\\10,&x=3\end{cases}\)?
▼ Show answerYes, the limit is \(5\) (the value \(f(3)=10\) is irrelevant)
Problem 5. Evaluate \(\displaystyle\lim_{x\to0^+}\sqrt{x}\).
▼ Show answer\(0\)
Problem 6. Find all values of \(k\) so that \(\displaystyle\lim_{x\to2}f(x)\) exists, where \(f(x)=\begin{cases}kx^2+1, & x<2\\ 3x+k, & x\ge2\end{cases}\).
▼ Show answer\(k=\dfrac53\)
Problem 7. Evaluate \(\displaystyle\lim_{x\to0}\dfrac{\sqrt{1+x}-\sqrt{1-x}}{x}\).
▼ Show answer\(1\)
Problem 8. The graph of \(y=f(x)\) is shown below. Find each of the following, or state that it does not exist: (a) \(\displaystyle\lim_{x\to1^-}f(x)\); (b) \(\displaystyle\lim_{x\to1^+}f(x)\); (c) \(\displaystyle\lim_{x\to1}f(x)\); (d) \(f(1)\); (e) \(\displaystyle\lim_{x\to3}f(x)\); (f) \(f(3)\).
▼ Show answer(a) \(2\); (b) \(3\); (c) does not exist, since the one-sided limits differ; (d) \(3\); (e) \(1\), since both sides approach the open circle at height \(1\); (f) \(2\) — the value \(f(3)\) is irrelevant to the limit in (e)
Computing Limits, Limit Laws, Squeeze Theorem [Lesson 5]
Limit Laws
For \(\lim f=L\), \(\lim g=M\): sum/difference \(L\pm M\); product \(LM\); quotient \(L/M\) (\(M\neq0\)); power \(L^n\); constant multiple \(cL\). For a polynomial, \(\lim_{x\to a}P(x)=P(a)\); for a rational function with \(Q(a)\neq0\), \(\lim_{x\to a}\frac{P(x)}{Q(x)}=\frac{P(a)}{Q(a)}\).
0/0 Indeterminate Forms
Factor and cancel the common factor, or rationalize (multiply by the conjugate) when a square root is involved, then substitute.
Squeeze Theorem
If \(f(x)\le g(x)\le h(x)\) for all \(x\) near \(a\) (except possibly at \(a\) itself) and \[\lim_{x\to a}f(x)=\lim_{x\to a}h(x)=L,\] then \(\displaystyle\lim_{x\to a}g(x)=L\) as well.
Practice Problems
Problem 1. Evaluate \(\displaystyle\lim_{x\to2}\dfrac{x^2-4}{x-2}\).
▼ Show answer\(4\)
Problem 2. Evaluate \(\displaystyle\lim_{x\to0}\dfrac{\sqrt{1+x}-1}{x}\).
▼ Show answer\(\dfrac12\)
Problem 3. Use the Squeeze Theorem to evaluate \(\displaystyle\lim_{x\to0}x^2\sin(1/x)\).
▼ Show answer\(0\)
Problem 4. Evaluate \(\displaystyle\lim_{x\to1}\dfrac{x^3-1}{x-1}\).
▼ Show answer\(3\)
Problem 5. Evaluate \(\displaystyle\lim_{x\to4}\dfrac{\sqrt{x}-2}{x-4}\).
▼ Show answer\(\dfrac14\)
Problem 6. Evaluate \(\displaystyle\lim_{x\to3}\dfrac{\dfrac1x-\dfrac13}{x-3}\).
▼ Show answer\(-\dfrac19\) (combine the fractions: \(\dfrac{3-x}{3x(x-3)}=-\dfrac{1}{3x}\), then substitute \(x=3\))
Problem 7. Evaluate \(\displaystyle\lim_{x\to2}\dfrac{x^3-8}{x^2-4}\).
▼ Show answer\(3\)
Infinite Limits & Vertical Asymptotes [Lesson 6]
Vertical Asymptotes
\(x=a\) is a vertical asymptote (VA) of \(f\) if any of \(\lim_{x\to a}f(x)\), \(\lim_{x\to a^+}f(x)\), or \(\lim_{x\to a^-}f(x)\) is infinite (\(+\infty\) or \(-\infty\)).
For a rational function \(f=p(x)/q(x)\) in lowest terms, vertical asymptotes occur exactly at the zeros of \(q(x)\). If a factor cancels between numerator and denominator, that point is a hole (removable), not a VA.
Common mistake: setting only the denominator to zero without checking whether the same factor cancels with the numerator.
Practice Problems
Problem 1. Find all vertical asymptotes of \(f(x) = \dfrac{x+2}{x^2-4}\).
▼ Show answer\(x=2\) only (there is a removable hole at \(x=-2\))
Problem 2. Evaluate \(\displaystyle\lim_{x\to3^+}\dfrac{1}{x-3}\).
▼ Show answer\(+\infty\)
Problem 3. Evaluate \(\displaystyle\lim_{x\to3^-}\dfrac{1}{x-3}\).
▼ Show answer\(-\infty\)
Problem 4. Find all vertical asymptotes of \(f(x) = \dfrac{x-1}{x^2-5x+6}\).
▼ Show answer\(x=2\) and \(x=3\)
Problem 5. Evaluate \(\displaystyle\lim_{x\to-2}\dfrac{x+3}{x^2-4}\), including one-sided behavior.
▼ Show answerDoes not exist; left-hand limit \(=+\infty\), right-hand limit \(=-\infty\)
Problem 6. Find the value of \(k\) so that \(f(x) = \dfrac{x^2+kx-6}{x-2}\) has a removable discontinuity (a hole) rather than a vertical asymptote at \(x=2\).
▼ Show answer\(k=1\)
Problem 7. For \(f(x) = \dfrac{2x}{x^2-x-6}\), find all vertical asymptotes and evaluate \(\displaystyle\lim_{x\to3^+}f(x)\).
▼ Show answerVertical asymptotes at \(x=3\) and \(x=-2\); \(\displaystyle\lim_{x\to3^+}f(x)=+\infty\)
Limits at Infinity & Horizontal/Slant Asymptotes [Lesson 7]
Limits at Infinity of Powers and Common Functions
A polynomial's end behavior is determined entirely by its leading term (the highest power dominates as \(x\to\pm\infty\)). Also useful: \(\lim_{x\to\infty}\tan^{-1}x=\pi/2\) and \(\lim_{x\to-\infty}\tan^{-1}x=-\pi/2\), so \(y=\pm\pi/2\) are horizontal asymptotes of \(\tan^{-1}x\).
End Behavior of Rational Functions
Compare degrees of \(p(x)\) and \(q(x)\) in \(f=p(x)/q(x)\):
- \(\deg p<\deg q\): \(\lim_{x\to\pm\infty}f(x)=0\) → horizontal asymptote \(y=0\)
- \(\deg p=\deg q\): limit \(=\) ratio of leading coefficients → that ratio is the horizontal asymptote
- \(\deg p>\deg q\): limit is \(\pm\infty\) — no horizontal asymptote
- \(\deg p=\deg q+1\): slant asymptote, found via long division
If \(f=p(x)/q(x)\) is in reduced form (no common factors), its vertical asymptotes are precisely the zeros of \(q(x)\) — see Lesson 6.
Square roots: for \(x>0\), \(\sqrt{x^2}=x\); for \(x<0\), \(\sqrt{x^2}=-x\) (i.e., \(\sqrt{x^2}=|x|\)). This sign flip matters when evaluating limits as \(x\to-\infty\) with a square root in the expression.
Practice Problems
Problem 1. Evaluate \(\displaystyle\lim_{x\to\infty}\dfrac{3x^2-5x}{2x^2+7}\).
▼ Show answer\(\dfrac32\)
Problem 2. Evaluate \(\displaystyle\lim_{x\to\infty}\dfrac{5x+2}{x^2-1}\).
▼ Show answer\(0\)
Problem 3. Evaluate \(\displaystyle\lim_{x\to-\infty}\dfrac{\sqrt{9x^2+1}}{x}\).
▼ Show answer\(-3\)
Problem 4. Find the slant asymptote of \(f(x) = \dfrac{x^2+3x-1}{x-2}\).
▼ Show answer\(y=x+5\)
Problem 5. Evaluate \(\displaystyle\lim_{x\to\infty}\dfrac{2x^3-1}{x^2+5}\).
▼ Show answer\(\infty\) (no horizontal asymptote)
Problem 6. Evaluate \(\displaystyle\lim_{x\to\infty}\left(\sqrt{x^2+3x}-x\right)\).
▼ Show answer\(\dfrac32\)
Problem 7. Evaluate \(\displaystyle\lim_{x\to-\infty}\dfrac{\sqrt{4x^2+x}}{3x-1}\).
▼ Show answer\(-\dfrac23\)
Continuity and Discontinuities [Lesson 8]
Definition of Continuity
\(f\) is continuous at \(x=a\) if and only if all the following three conditions are satisfied:
(1) \(f(a)\) is defined;
(1) (2) \(\lim_{x\to a}f(x)\) exists (and is finite);
(1) (3) \(\lim_{x\to a}f(x)=f(a)\)
Polynomials, rational functions (away from zeros of the denominator), compositions of continuous functions, inverse functions, and inverse trig functions are all continuous on their domains; a piecewise function is continuous wherever each piece is continuous and the pieces agree at the break points.
Types of Discontinuity
- Removable: \(\lim_{x\to a}f(x)\) exists, but \(f(a)\) is either undefined or not equal to the limit
- Jump: left and right limits exist but differ
- Infinite: a one-sided limit is \(\pm\infty\) (a vertical asymptote)
Limits of Composite Functions
(a) If \(g\) is continuous at \(a\) and \(f\) is continuous at \(g(a)\), then \[\lim_{x\to a}f(g(x)) = f\!\left(\lim_{x\to a}g(x)\right) = f(g(a)).\]
(b) More generally, if \(\lim_{x\to a}g(x)=L\) and \(f\) is continuous at \(L\), then \[\lim_{x\to a}f(g(x)) = f\!\left(\lim_{x\to a}g(x)\right) = f(L).\]
Intermediate Value Theorem (IVT)
If \(f\) is continuous on \([a,b]\) and \(L\) is any value between \(f(a)\) and \(f(b)\), then there is at least one \(c\in(a,b)\) with \(f(c)=L\).
Practice Problems
Problem 1. Find \(k\) so that \(f(x)=\begin{cases}x^2-1,&x<3\\kx+2,&x\ge3\end{cases}\) is continuous at \(x=3\).
▼ Show answer\(k=2\)
Problem 2. Classify the discontinuity of \(f(x) = \dfrac{x^2-9}{x-3}\) at \(x=3\).
▼ Show answerRemovable (a hole); the limit is \(6\) but \(f(3)\) is undefined
Problem 3. Where is \(f(x) = \dfrac{1}{x^2-x-6}\) discontinuous?
▼ Show answer\(x=3\) and \(x=-2\) (infinite discontinuities)
Problem 4. Use the IVT to show that \(f(x)=x^3-x-1\) has a root in \((1,2)\).
▼ Show answer\(f(1)=-1<0\) and \(f(2)=5>0\); since \(f\) is continuous, IVT guarantees a root in \((1,2)\)
Problem 5. Is \(f(x) = \dfrac{|x-2|}{x-2}\) continuous at \(x=2\)?
▼ Show answerNo; \(f\) is undefined at \(x=2\) and the one-sided limits differ (\(-1\) vs. \(1\))
Problem 6. Find \(a\) and \(b\) so that \(f(x)=\begin{cases}x^2+a, & x<1\\ b, & x=1\\ 2x+3, & x>1\end{cases}\) is continuous at \(x=1\).
▼ Show answer\(a=4\), \(b=5\)
Problem 7. Use the IVT to show that \(\cos x = x\) has a solution in \((0,\pi/2)\).
▼ Show answerLet \(f(x)=\cos x-x\). \(f(0)=1>0\) and \(f(\pi/2)=-\pi/2<0\); since \(f\) is continuous, IVT guarantees a root, i.e., a solution to \(\cos x=x\), in \((0,\pi/2)\)
Problem 8. The graph of \(y=f(x)\) is shown below. (a) At which \(x\)-values is \(f\) discontinuous? Classify each discontinuity as removable, jump, or infinite. (b) At which of those \(x\)-values does \(\displaystyle\lim f(x)\) exist?
▼ Show answer(a) \(x=-2\): removable — the limit is \(1\) but \(f(-2)\) is a different value (about \(2.5\)); \(x=1\): jump — the left-hand limit (about \(2.5\)) and right-hand limit (about \(0.5\)) differ; \(x=3\): infinite — a vertical asymptote. (b) Only at \(x=-2\)
Introduction to Derivatives [Lesson 9]
Derivative at a Point
Three interpretations of \(f'(a)\): the slope of the tangent line to \(y=f(x)\) at \(x=a\); the (instantaneous) velocity at time \(a\), if \(f\) is a position function; and, in general, the instantaneous rate of change of \(f(x)\) at \(a\).
Tangent Line and Normal Line
Tangent line at \(a\) (point-slope form, slope \(m=f'(a)\)): \[y-f(a)=f'(a)(x-a)\]
Normal line at \(a\) (perpendicular to the tangent, when \(f'(a)\neq0\)): \[y-f(a)=-\frac{1}{f'(a)}(x-a)\]
The normal line is perpendicular to the tangent line at the same point; if the tangent has slope \(m\), the normal has slope \(-1/m\).
Practice Problems
Problem 1. Use the limit definition to find \(f'(2)\) for \(f(x)=x^2+3x\).
▼ Show answer\(f'(2)=7\)
Problem 2. Find the equation of the tangent line to \(f(x)=x^2\) at \(x=1\).
▼ Show answer\(y=2x-1\)
Problem 3. Evaluate \(\displaystyle\lim_{h\to0}\dfrac{(3+h)^2-9}{h}\) and identify what derivative this represents.
▼ Show answerThis is \(f'(3)\) for \(f(x)=x^2\); the limit equals \(6\)
Problem 4. If \(f(x)=\sqrt{x}\), use the limit definition to find \(f'(4)\).
▼ Show answer\(f'(4)=\dfrac14\)
Problem 5. Determine whether \(f(x)=|x|\) is differentiable at \(x=0\).
▼ Show answerNo; the left-hand derivative is \(-1\) and the right-hand derivative is \(1\), so they disagree
Problem 6. Use the limit definition to find a general formula for \(f'(x)\) if \(f(x) = \dfrac{1}{x+1}\).
▼ Show answer\(f'(x) = -\dfrac{1}{(x+1)^2}\)
Problem 7. Find all points on the graph of \(f(x)=x^3-3x\) where the tangent line is horizontal.
▼ Show answer\((1,-2)\) and \((-1,2)\)
The Derivative as a Function [Lesson 10]
Definition
This is a new function giving the tangent slope at every \(x\). A function is differentiable at \(x\) if this limit exists there.
Key implication: if \(f\) is differentiable at \(x=a\), then \(f\) is continuous at \(a\). The converse is false — a function can be continuous at a point without being differentiable there (e.g. \(f(x)=|x|\) at \(x=0\)).
When \(f'(a)\) Fails to Exist
\(f'(a)\) fails to exist when there is a vertical tangent at \(a\); \(f\) is discontinuous at \(a\); or there is a corner/cusp at \(a\) (left- and right-hand derivatives disagree, as with \(f(x)=|x|\) at \(x=0\)).
Graphical Relationship
Where \(f\) is increasing, \(f'>0\); where \(f\) is decreasing, \(f'<0\); where \(f\) has a horizontal tangent, \(f'=0\).
Practice Problems
Problem 1. Given \(f(x)=x^3-3x\), find \(f'(x)\) using the limit definition and evaluate \(f'(1)\).
▼ Show answer\(f'(x)=3x^2-3\); \(f'(1)=0\)
Problem 2. At which \(x\)-value does \(f(x)=|x-3|\) fail to be differentiable?
▼ Show answer\(x=3\) (a corner)
Problem 3. If \(f\) is increasing on \((a,b)\), what can be said about \(f'(x)\) there?
▼ Show answer\(f'(x)\ge0\) on \((a,b)\) (and \(>0\) wherever \(f\) is strictly increasing and differentiable)
Problem 4. Use the limit definition to find \(f'(x)\) for \(f(x)=\sqrt{2x+1}\), and state the domain of \(f'\).
▼ Show answer\(f'(x)=\dfrac{1}{\sqrt{2x+1}}\) (multiply by the conjugate); domain of \(f'\) is \((-\tfrac12,\infty)\) — the endpoint \(x=-\tfrac12\) of the domain of \(f\) is excluded
Problem 5. Explain why \(f(x)=x^{2/3}\) is not differentiable at \(x=0\).
▼ Show answer\(f'(x)=\tfrac23x^{-1/3}\) is unbounded near \(0\): it tends to \(+\infty\) as \(x\to0^+\) and to \(-\infty\) as \(x\to0^-\), so the limit defining \(f'(0)\) does not exist. The graph has a cusp with a vertical tangent at the origin
Problem 6. Let \(f(x)=\begin{cases}x^2, & x\le1\\ 2x-1, & x>1\end{cases}\). Is \(f\) differentiable at \(x=1\)? Justify using one-sided derivatives.
▼ Show answerYes. \(f\) is continuous at \(1\) (both pieces give \(1\)); the left-hand derivative is \(2x\big|_{x=1}=2\) and the right-hand derivative is \(2\), so they agree and \(f'(1)=2\)
Problem 7. Find \(a\) and \(b\) so that \(f(x)=\begin{cases}ax^2+b, & x<1\\ 3x, & x\ge1\end{cases}\) is differentiable at \(x=1\).
▼ Show answer\(a=\dfrac32\), \(b=\dfrac32\). Differentiability requires continuity (\(a+b=3\)) and matching one-sided derivatives (\(2a=3\))
Problem 8. The graph of \(y=f(x)\) on \([-3,5.5]\) is shown below. (a) On which intervals is \(f'(x)>0\)? (b) At which \(x\)-values is \(f'(x)=0\)? (c) At which \(x\)-value does \(f'(x)\) fail to exist, and why? (d) Is \(f'(0)\) positive or negative?
▼ Show answer(a) \((-3,-1)\) and \((2,4)\), where \(f\) is increasing; (b) \(x=-1\) and \(x=2\), where the tangent line is horizontal; (c) \(x=4\): the graph has a corner, so the left- and right-hand slopes disagree; (d) negative, since \(f\) is decreasing at \(x=0\)