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Comprehensive Calculus II Exam Prep: Sketching, Continuity, Tangent Lines, Limit-Based Derivatives, and Rule Applications

  • Sketching from given properties (explicit sketching): you’re given a set of function properties (points to hit, limits at certain x, end behavior, asymptotes) and you must sketch any function that satisfies them. - Key checks: verify the vertical test (the function must pass the vertical line test to be a function); ensure behavior matches specified limits and asymptotes; identify where the function is continuous vs. discontinuous; recognize removable discontinuities (limit exists but value not equal to the limit). - Example scenario described: given a few x-values and required y-values, plus a specified limit at a point and a vertical asymptote somewhere. One sketch shown had: a point the graph must pass through (e.g., (2,3)); a left/right limit at that x equal to a specified value (e.g., a limit of 2); a vertical asymptote elsewhere; and ends rising to +∞ on one or both sides. There can be infinitely many functions meeting these conditions, so the goal is to satisfy all constraints with a coherent picture. - Another point: sometimes you’re allowed to specify only one-sided limits for a condition; providing both one-sided limits narrows your options, but you can still meet the given criteria.
  • Continuity and differentiability in piecewise and constructed examples: continuity at a point means the left-hand limit equals the right-hand limit and equals the function value. Differentiability requires the left- and right-hand derivatives to exist and match. - Example outline discussed: a piecewise function with potential nondifferentiability at breakpoints (e.g., at x=0 and x=1). In the example, the left-hand and right-hand limits at 0 both equal 1, so the function is continuous at 0, but the one-sided derivative values differ, so the function is not differentiable at 0. At x=1, the left-hand and right-hand derivatives did not match, so the function is not differentiable at x=1. - A one-small-change note from the discussion: changing the middle piece’s sign (e.g., from a minus to a plus) can alter continuity/differentiability at a breakpoint (e.g., making the function differentiable at x=0 in the adjusted setup).
  • Tangent lines and horizontal tangents:
    • Tangent line formula: y - y1 = m (x - x1) where m = f'(x1) and the point is $(x1, f(x_1))$.
    • Horizontal tangent lines occur when f'(x) = 0 ; equivalently, find where the derivative vanishes.
    • Example discussed: for a simple function (e.g., a cubic like f(x) = rac{1}{3}x^3 ), the tangent line at a chosen point can be found by computing the derivative and substituting the point. If we pick a clean choice, the tangent line can be written in point-slope form and then simplified to a slope-intercept form.
    • Another example topic: solving for horizontal tangents yielded zeros of the derivative, e.g., if f'(x) = x^2 - 5x + 6 = (x-2)(x-3) then x = 2 ext{ or } x = 3 are the x-coordinates of horizontal tangents.
  • Derivative from the limit definition (three common types) and a worked limit example:
    • Definition: f'(a) = oxed{ ext{lim}_{h o 0} rac{f(a+h)-f(a)}{h}}.
    • Three common types of derivative-limit problems:
    • Polynomials: expand the numerator, simplify, cancel a factor of $h$.
    • Rational expressions: put over a common denominator to cancel $h$.
    • Square-root expressions: multiply by the conjugate to create a difference-of-squares and cancel $h$.
    • Worked example (square root): compute the derivative of f(x)=
      oot 2 o ext{sqrt}{2x+1} via the limit definition:
    • Start with f'(x)= ext{lim}_{h o 0} rac{
      oot 2 o ext{sqrt}{2(x+h)+1}-
      oot 2 o ext{sqrt}{2x+1}}{h}.
    • Multiply numerator and denominator by the conjugate
      oot 2 o ext{sqrt}{2(x+h)+1} +
      oot 2 o ext{sqrt}{2x+1} to get
      f'(x) = ext{lim}{h o 0} rac{(2x+2h+1)-(2x+1)}{h ig( oot 2 o ext{sqrt}{2(x+h)+1} + oot 2 o ext{sqrt}{2x+1}ig)} = ext{lim}{h o 0} rac{2h}{h ig(
      oot 2 o ext{sqrt}{2(x+h)+1} +
      oot 2 o ext{sqrt}{2x+1}ig)}.
    • Cancel $h$ and take the limit as $h o 0$ to get
      f'(x) = rac{2}{
      oot 2 o ext{sqrt}{2x+1}+
      oot 2 o ext{sqrt}{2x+1}} = rac{2}{2
      oot 2 o ext{sqrt}{2x+1}} = rac{1}{
      oot 2 o ext{sqrt}{2x+1}}.
    • Alt approach via the chain rule confirms the same result: f'(x)= rac{1}{
      oot 2 o ext{sqrt}{2x+1}}.
    • Chain rule and nested functions: outer function first, then inside, and so on. Example: f(x)= rac{}{} ext{sin}( ext{sec}(x^5))
    • Let $u= ext{sec}(x^5)$, then $f(x)= ext{sin}(u)$; $f'(x)= ext{cos}(u) rac{du}{dx}$. Now $ rac{du}{dx}= ext{sec}(x^5) an(x^5) rac{d}{dx}(x^5)= ext{sec}(x^5) an(x^5)ig(5x^4ig)$. Thus
      f'(x)= ext{cos}( ext{sec}(x^5))ig[ ext{sec}(x^5) an(x^5)ig] 5x^4.
    • If you have a product, apply the product rule: $(gh)'=g'h+gh'$; if you have a quotient, apply the quotient rule: $(u/v)'= rac{u'v-uv'}{v^2}$. If a chain rule is involved, apply it to the inner and outer functions.
    • An alternative framing used in class: you can treat a quotient as a product with an exponent $-1$ (e.g., rac{u(x)}{v(x)} = u(x)ig(v(x)^{-1}ig) ) when convenient for differentiation.
    • Practical exam guidance: show enough work to justify the derivative; it’s acceptable to leave final answers unsimplified if justified; for product/chain/quotient problems you’ll likely need to show the rule you used and the key intermediate derivatives.
  • Absolute-value limits (one-sided analysis):
    • When limits involve absolute values, determine the sign of the inside near the limit point on each side, then remove the absolute value accordingly to compute one-sided limits.
    • Example (two-sided limit that does not exist): compute
      oxed{ ext{lim}_{x o 4} rac{|2x-8|}{x-4} }
    • For $x<4$, $|2x-8|=-(2x-8)=8-2x$, so
      ext{lim}{x o 4^-} rac{8-2x}{x-4} = ext{lim}{x o 4^-} rac{-2(x-4)}{x-4} = -2.
    • For $x>4$, $|2x-8|=2x-8$, so
      ext{lim}{x o 4^+} rac{2x-8}{x-4} = ext{lim}{x o 4^+} rac{2(x-4)}{x-4} = 2.
    • Since the left and right limits differ, the two-sided limit does not exist. If the absolute value were absent, you’d just have a simple linear limit; the absolute value can create a jump in the limit from each side.
  • Trigonometry basics for the exam:
    • You should be able to evaluate trig functions at common angles; focus on quadrant I basics and standard angles.
    • Key values to know (in radians):
    • heta = 0, rac{
      ho}{6}, rac{
      ho}{4}, rac{
      ho}{3}, rac{
      ho}{2} where
      ho = ext{pi}, i.e., 0, $ rac{ ext{pi}}{6}$, $ rac{ ext{pi}}{4}$, $ rac{ ext{pi}}{3}$, $ rac{ ext{pi}}{2}$.
    • Fundamental values: egin{aligned} ext{sin}(0)&=0,\n ext{cos}(0)&=1,\
      ext{sin}( frac{ ext{pi}}{6})&= frac{1}{2},\ ext{cos}( frac{ ext{pi}}{6})&= frac{
      oot 3 o ext{3}}{2},\ ext{sin}( frac{ ext{pi}}{4})&= frac{
      oot 2 o ext{2}}{2},\ ext{cos}( frac{ ext{pi}}{4})&= frac{
      oot 2 o ext{2}}{2},\ ext{sin}( frac{ ext{pi}}{3})&= frac{
      oot 3 o ext{3}}{2},\ ext{cos}( frac{ ext{pi}}{3})&= frac{1}{2},\ ext{sin}( frac{ ext{pi}}{2})&=1, ext{, } ext{cos}( frac{ ext{pi}}{2})=0. \ ext{These values, together with reciprocal identities, can yield all needed values.}

ight.

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  • Exam logistics and study tips (course context):
    • Expect a short, exam-style review day format with concise true/false questions (two points each; explanation earns points). One or two problems may involve the derivative as a limit definition. A typical problem set contains about 10 problems.
    • Types likely on the exam include: a derivative via the limit definition (with polynomials, fractions, or square roots via conjugates), a function having horizontal tangents, a tangent-line problem, a piecewise-continuity/differentiability check, and a limit involving an absolute value.
    • Helpful study practices: review position-velocity-acceleration relationships, horizontal tangent methods, how to identify differentiability points from left and right limits, and how to craft clear justification for true/false questions.
    • Practical exam-day tips: show work that demonstrates understanding (even partial credit if your final answer isn’t perfect), and be prepared to justify steps (you can ask clarifying questions if you’re unsure about how much work to show).
  • Final logistics mentioned in the session:
    • Exams and skills tests scheduling: a skills test later in the week and another next week; quiz timing may be adjusted to avoid a heavy last week; office hours available for review after class.
    • Instructor emphasizes accessibility of grading and that you can check grade details during office hours. If you email for studying help, expect a delay on Fridays.
  • Summary of key exam-ready concepts to focus on:
    • Explicit sketching from given properties: passes through specified points, includes limits and asymptotes, and demonstrates possible nondifferentiability (cusp) or discontinuities.
    • Continuity test: left-hand limit = right-hand limit = f(c) for continuity at x=c.
    • Differentiability test: continuity plus equality of left- and right-hand derivatives at the point.
    • Tangent lines and horizontal tangents: tangent line equation, slope from derivative value at the point, and solving f'(x)=0 for horizontal tangents.
    • Derivative from limit definition: correct use of the limit, with appropriate algebra for polynomials, fractions, and square roots (including conjugates for square roots).
    • Chain rule, product rule, and quotient rule: correct application order and when to use each; understanding that chain rule is applied from the outermost function inward.
    • Absolute value limits: determine one-sided limits by removing the absolute value according to the sign on each side, and recognize when the two-sided limit does not exist.
    • Trig value evaluation: know standard values for common angles and how to extend with identities.
  • Quick worked-check hints (optional):
    • For a derivative via limit with a square root: check the conjugate step carefully to ensure the difference of squares is used correctly.
    • For a chain-rule-heavy function like f(x)= ext{sin}( ext{sec}(x^5)) , keep track of the inside function derivatives in steps; the outermost derivative is applied first (chain rule).
    • When presenting tangent lines, you can present the equation in point-slope form and then reorganize to slope-intercept form if requested, but always cite the slope as the derivative at the point.
  • Note on instructor's examples: some arithmetic in the transcript contains minor numerical inconsistencies (e.g., mismatched derivatives in an example using f(x)=(1/3)x^3$$ or in an intermediate product-rule computation). The correct results are provided above where applicable; use the correct derivative rules when studying and practice verifying each step on your own.