Continuity (College Board AP® Calculus BC): Exam Questions

The graph of a function $f$ is shown on the diagram above. Which of the following statements about $f$ is false?

$f\left(x\right)=\{\begin{array}{l}\frac{\left(3x+4\right)\left(x+1\right)}{x+1}\mathrm{for}x\ne -1\\ k\mathrm{for}x=-1\end{array}$

Let $f$ be the function defined above. For what value of $k$ is $f$ continuous at $x=-1$?

Let $f$ be a function that is continuous on the closed interval $\left[3,7\right]$ with $f\left(3\right)=12$ and $f\left(7\right)=18$. Which of the following is guaranteed by the Intermediate Value Theorem?

The function $f$ is continuous on the closed interval $\left[-1,1\right]$ and has values that are given in the table above. The equation $f\left(x\right)=8$ must have at least two solutions in the interval $\left[-1,1\right]$ if k$=$

$f\left(x\right)=\{\begin{array}{l}3\mathrm{for}x\le a\\ \frac{1}{x-a}\mathrm{for}x>a\end{array}$

Let $f$ be the function defined above. Which of the following statements is true?

$f\left(x\right)=\{\begin{array}{l}2x^2-3x+1\mathrm{for}x\le a\\ x^2+3x-8\mathrm{for}x>a\end{array}$

The function $f$ defined above is a continuous function. What is the value of $a$?

$f\left(x\right)=\{\begin{array}{l}\frac{x^2-7x+10}{x^2+5x-14}\mathrm{for}x\ge -5,x\ne 2\\ k\mathrm{for}x=2\end{array}$

Let $f$ be the function defined above. For what value of $k$ is $f$ continuous at $x=2$?

$f\left(x\right)=\{\begin{array}{l}{\left(x-a\right)}^2\mathrm{for}x\le a\\ \cos \left(\frac{1}{x^2-a^2}\right)\mathrm{for}x>a\end{array}$

Let $f$ be the function defined above. Which of the following statements is true?

Let $f$ be a function that is differentiable on the open interval $\left(0,11\right)$. If $f\left(1\right)=4$, $f\left(7\right)=-3$, and $f\left(10\right)=9$, which of the following must be true?

I.  $f$ has at least 2 zeros.

II.  The graph of $f$ has at least one horizontal tangent.

III.  For some $c$, $7<c<10$, $f\left(c\right)=-2$.

Let $f$ be the function defined by $f\left(x\right)=\tan x$. To 3 decimal places, $f\left(1\right)=1.557$ and $f\left(2\right)=-2.185$. A student claims that, according to the intermediate value theorem, there is some value $c$ between 1 and 2 such that $f\left(c\right)=0$.

Why is the student's claim not valid?