AP Calculus AB · Free-response skill

Justify continuity and use the IVT

The move. Show that a function is continuous at a point by checking all three conditions with values, then use continuity to argue with the Intermediate Value Theorem that a value is reached.

How to do it

  1. Continuity at a point is three checks, and a full answer shows all three with numbers: the value f(a)f(a) exists, the limit exists (show that the left and right limits agree), and the limit equals the value.
  2. For a piecewise function, find each one-sided limit from the piece that holds on that side. If a piece gives 0 over 0, simplify it before you substitute.
  3. To use the Intermediate Value Theorem, write three things: the function is continuous on the closed interval (and why), the target value lies between the two endpoint values (give both values), and the theorem by name with its conclusion.
  4. The theorem promises at least one input with the target value. It does not say where, and it promises nothing when the target is not between the endpoint values.
  5. No step of this question needs a calculator, so practice it fully by hand.

Practice prompt Try it

Let gg be the function defined by g(x)=x2−4x−2g(x) = \frac{x^2 - 4}{x - 2} for x<2x < 2 and g(x)=h(x)g(x) = h(x) for x≥2x \ge 2, where hh is a function that is continuous for all x≥2x \ge 2. Selected values of hh are shown in the table.

Selected values of hh

xx2357
h(x)h(x)491-3

Use the definition of gg and the table to answer each part.

  1. Find lim⁡x→2−g(x)\lim_{x \to 2^-} g(x). Show the algebra you use.
  2. Decide whether gg is continuous at x=2x = 2. Check each of the three conditions for continuity, using values.
  3. Must there be a value of xx between 0 and 3 at which g(x)=6g(x) = 6? Give a reason, naming the theorem you use and saying why its condition holds.

Write each part, then check yourself against the checklist. Nothing you type here is saved or graded.

Check yourself and see a model answer

Checklist

Our own checklist of what a full answer does, one line per point. Tick each one your response earns.

0 of 7 points

Model answer, annotated

(a) For x<2x < 2, substitution gives 00\frac{0}{0}, so factor: g(x)=(x−2)(x+2)x−2=x+2g(x) = \frac{(x - 2)(x + 2)}{x - 2} = x + 2 for x<2x < 2. So lim⁡x→2−g(x)=lim⁡x→2−(x+2)=4\lim_{x \to 2^-} g(x) = \lim_{x \to 2^-}(x + 2) = 4.

  • Point 1 Factors and cancels x - 2 to remove the 0 over 0 form, then substitutes to get 4.

(b) First, g(2)=h(2)=4g(2) = h(2) = 4 from the table, so g(2)g(2) is defined. Second, for x>2x > 2, g(x)=h(x)g(x) = h(x), and hh is continuous for x≥2x \ge 2, so lim⁡x→2+g(x)=lim⁡x→2+h(x)=h(2)=4\lim_{x \to 2^+} g(x) = \lim_{x \to 2^+} h(x) = h(2) = 4. With part (a), both one-sided limits are 4, so lim⁡x→2g(x)=4\lim_{x \to 2} g(x) = 4 exists. Third, this limit equals g(2)=4g(2) = 4. All three conditions hold, so gg is continuous at x=2x = 2.

  • Point 2 Reads g(2) = h(2) = 4 from the table: the value exists.
  • Point 3 Uses the continuity of h for x at least 2 to get the right-hand limit 4.
  • Point 4 Puts the matching one-sided limits together and compares with g(2) to conclude continuity.

(c) Yes. On [0,2)[0, 2), g(x)=x2−4x−2g(x) = \frac{x^2 - 4}{x - 2} is a rational function whose denominator is not 0, so it is continuous there; gg is continuous at 2 by part (b); and on [2,3][2, 3], g=hg = h is continuous. So gg is continuous on [0,3][0, 3]. Next, g(0)=0−40−2=2g(0) = \frac{0 - 4}{0 - 2} = 2 and g(3)=h(3)=9g(3) = h(3) = 9, and 6 is between 2 and 9. By the Intermediate Value Theorem, g(c)=6g(c) = 6 for some cc with 0<c<30 < c < 3.

  • Point 5 Explains why g is continuous on all of [0, 3], piece by piece and at the breakpoint.
  • Point 6 Computes both endpoint values and places 6 between them.
  • Point 7 Names the theorem and states its conclusion for this interval.