AP Calculus AB Continuity and the Intermediate Value Theorem Quiz: 10 Questions with Explanations

A jumpB removableC infiniteD none; it is continuous

The limit is 3 but f(2) = 1, so a hole with a misplaced point is a removable discontinuity.

A removableB jumpC infiniteD none; it is continuous

The one-sided limits (2 and 5) exist but differ, which is a jump.

A removableB jumpC infiniteD oscillating

The outputs grow without bound near x = 2, so there is a vertical asymptote (infinite discontinuity).

A \(f\) is differentiable at \(a\)B \(f(a)\) is definedC \(\lim_{x\to a} f(x)\) existsD \(\lim_{x\to a} f(x)=f(a)\)

Continuity needs only the three limit conditions. Differentiability is a stronger property that is not required.

A \(1\)B \(3\)C \(\frac{4}{3}\)D \(2\)

Match values at 3: 4 = 3k – 2 gives k = 2. The value 4/3 comes from dropping the -2 and solving 3k = 4.

A \(k=7\)B \(k=3\)C \(k=-3\)D \(k=10\)

Only 3 lies between -1 and 6. The other values are outside the range of the endpoint values.

A \(f(x)=x^{2}+1\)B \(f(x)=\frac{1}{x^{2}-1}\)C \(f(x)=\frac{x^{2}-1}{x-1}\)D \(f(x)=\frac{|x|}{x}\)

A polynomial is continuous everywhere. The others have a zero denominator at x = 1, x = 1 or x = 0.

A \(f(0)=-1\) and \(f(1)=-1\)B \(f(1)=-1\) and \(f(3)=23\)C \(f(2)=5\) onlyD \(f(1)=-1\) and \(f(2)=5\)

A sign change on [1, 2] proves a root there: f(1) = -1 is negative and f(2) = 5 is positive. Other intervals do not bracket the root or have no sign change.

A \(f(-1)\) is negativeB \(f\) is a rational functionC \(f\) is not continuous at \(x=0\)D the interval is closed

The theorem needs continuity on the entire interval. The asymptote at x = 0 breaks that.

A it has a jump discontinuity at \(x=2\)B it has a removable discontinuity at \(x=2\)C it is continuous at \(x=2\)D it has a vertical asymptote at \(x=2\)

Different finite one-sided limits mean a jump. The limit does not exist, so it cannot be removable or continuous.