Question
$f(x)= \begin{cases}3 e^{x-1}-2 & \text { for } x \leq 1 \\ 4 x^2+b x-5 & \text { for } x>1\end{cases}$
Consider the function $f(x)$, defined above, where $b$ is a constant. What is the value of $b$ for which the function $f$ is continuous at $x=1 ?$
(A) $-9$
(B) $-4$
(C) 1
(D) 2
▶️Answer/Explanation
Ans:D
Since we want the function continuous at $x=1$, then $\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(1)$. First find the two one-sided limits, and then solve for $b$.
Left limit: $\lim _{x \rightarrow 1^{-}} f(x)=3 e^{1-1}-2=1$
Right limit: $\lim _{x \rightarrow 1^{+}} f(x)=4(1)^3+b \cdot 1-5=b-1$
Set the one-sided limits equal and solve for $b: b-1=1 \Rightarrow b=2$. Therefore, since $f(1)=1$, when $b=2$, both onesided limits are equal and are equal to $f(1)$.
Question
Function $h$ is twice differentiable. The table above gives selected values of $h$. Which of the following must be true?
(A) $h$ has no critical points in the interval $-4<x<2$.
(B) $h^{\prime}(x)=10$ for some value of $x$ in the interval $-4<x<2$.
(C) The graph of $h$ has no points of inflection in the interval $-4<x<2$.
(D) $h^{\prime}(x)>0$ for all values of $x$ in the interval $-4<x<2$.
▶️Answer/Explanation
Ans:B
Since the function $h$ is twice differentiate, it is continuous on the interval $[-4,2]$ so the Mean Value Theorem applies. Therefore, there must be a value of $x$ in the interval $(-4,2)$
such that $h^{\prime}(x)=\frac{h(2)-h(-4)}{2-(-4)}=\frac{24-(-36)}{6}=\frac{60}{6}=10$.
NOTE: $h$ must have a critical point since the Intermediate Value Theorem applies on $h^{\prime}(x)$. There is no way to determine if there is an inflection point since we don’t know the behavior of $h$ between the selected $x$-values in the table. Although the values of $h$ that we see on the interval $[-4,2]$ are increasing, we don’t know the function values between these selected values of $x$. So $h$ could be decreasing on some interval.
Question
If $x=\sqrt{1-t^2}$ and $y=\sin ^{-1} t$, then $\frac{d y}{d x}$ equals
(A) $-\frac{\sqrt{1-t^2}}{t}$
(B) $-t$
(C) 2
(D) $-\frac{1}{t}$
▶️Answer/Explanation
Ans:D
Here,
$
\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{1}{\sqrt{1-t^2}}}{\frac{1}{2} \frac{(-2 t)}{\sqrt{1-t^2}}}=-\frac{1}{t}
$
Question
If $\sum_{n=1}^{\infty} b_n$ is a geometric series of all-positive terms with $b_1=90$ and $b_3=10$, then $\sum_{n=1}^{\infty} b_n$
(A) diverges
(B) $=105$
(C) $=135$
(D) converges to a sum that cannot be determined
▶️Answer/Explanation
Ans:C
If $b_1=90$ and $b_3=10$, this geometric series has a common ratio of $\frac{1}{3}$. Thus the sum of the series is $\frac{90}{1-\frac{1}{3}}=90 \cdot \frac{3}{2}=135$. Note that the sum cannot be 105 because the sum of the first and third terms is already 100 and the second term must be greater than 10 .
Question
The table above shows values of differentiable functions $f$ and $g$. If $h(x)=g(f(x))$ then $h^{\prime}(3)=$
(A) $\frac{1}{2}$
(B) 1
(C) 4
(D) 6
▶️Answer/Explanation
Ans:B