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Continuity and Discontinuities at x = 2

Created by @nathanedwards
 at November 1st 2023, 7:02:25 pm.
AP_Calculus_AB-Questions
#calculus
#continuity
#discontinuity

AP Calculus AB Exam Question: Continuity and Types of Discontinuities

Consider the function given by:

f(x)={x2−4x−2x<2kx=25x+10x−6x>2 f(x) = \begin{cases} \frac{x^2-4}{x-2} & x<2 \\ k & x=2 \\ \frac{5x+10}{x-6} & x>2 \end{cases}
  1. Determine the value of k that would make the function continuous at x = 2.
  2. Classify the type(s) of discontinuity (if any) at x = 2 for values of k that do not satisfy the condition for continuity.

Answer:

  1. To find the value of k that makes the function continuous at x = 2, we need to ensure that the limit of the function as x approaches 2 from both the left and the right exists and is equal to the value of the function at x = 2.

Step 1: Calculate the left-hand limit:

lim⁡x→2−x2−4x−2 \lim_{{x \to 2^-}} \frac{x^2-4}{x-2}

Applying direct substitution, we get:

lim⁡x→2−22−42−2=lim⁡x→2−4−40=lim⁡x→2−00 \lim_{{x \to 2^-}} \frac{2^2-4}{2-2} = \lim_{{x \to 2^-}} \frac{4-4}{0} = \lim_{{x \to 2^-}} \frac{0}{0}

At this point, we have an indeterminate form, so we can try factoring the numerator:

lim⁡x→2−(x−2)(x+2)x−2 \lim_{{x \to 2^-}} \frac{(x-2)(x+2)}{x-2}

Now, we can cancel out the common factor of (x-2):

lim⁡x→2−(x+2)=2+2=4 \lim_{{x \to 2^-}} (x+2) = 2+2 = 4

Step 2: Calculate the right-hand limit:

lim⁡x→2+5x+10x−6 \lim_{{x \to 2^+}} \frac{5x+10}{x-6}

Applying direct substitution, we get:

lim⁡x→2+5(2)+102−6=lim⁡x→2+20−4=−5 \lim_{{x \to 2^+}} \frac{5(2)+10}{2-6} = \lim_{{x \to 2^+}} \frac{20}{-4} = -5

Step 3: Set up the equation for continuity and solve for k:

lim⁡x→2−f(x)=lim⁡x→2+f(x)=f(2) \lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^+}} f(x) = f(2)

Plugging in the limits and the value of k, we have:

4=−5=k 4 = -5 = k

Therefore, the value of k that makes the function continuous at x = 2 is -5.

  1. For values of k that do not satisfy the condition for continuity, there will be a discontinuity at x = 2. We need to determine the type(s) of discontinuity.

From our calculations in step 1 and step 2, we found that the left-hand limit is 4 and the right-hand limit is -5. Since these two limits are not equal, we have a jump discontinuity at x = 2 for values of k that do not satisfy the condition for continuity.

Thus, depending on the value of k, the function may have a jump discontinuity at x = 2.

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