Home / AP Calculus BC: 1.15 Connecting Limits at Infinity and Horizontal Asymptotes- Exam Style questions with Answer- FRQ

AP Calculus BC: 1.15 Connecting Limits at Infinity and Horizontal Asymptotes- Exam Style questions with Answer- FRQ

Question

(a) Topic-1.15 Connecting Limits at Infinity and Horizontal Asymptotes

(b) Topic-5.5 Using the Candidates Test to Determine Absolute (Global) Extrema

(c) Topic-5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

(d) Topic-5.11 Solving Optimization Problems

Let $f$ be the function given by $f(x) = 2xe^{2x}$.
(a) Find $\lim_{x \to -\infty} f(x)$ and $\lim_{x \to \infty} f(x)$.
(b) Find the absolute minimum value of $f$. Justify that your answer is an absolute minimum.
(c) What is the range of $f$?
(d) Consider the family of functions defined by $y = bxe^{bx}$, where $b$ is a nonzero constant. Show that the absolute minimum value of $bxe^{bx}$ is the same for all nonzero values of $b$.

▶️Answer/Explanation

\(\textbf{2(a)}\)
$\lim_{x \to -\infty} 2xe^{2x} = 0$
$\lim_{x \to \infty} 2xe^{2x} = \infty$ or DNE

\(\textbf{2(b)}\)
$f'(x) = 2e^{2x} + 2x \cdot 2 \cdot e^{2x} = 2e^{2x}(1 + 2x) = 0$
If $x = -1/2$, then $f(-1/2) = -1/e$ or approximately $-0.368$ or $-0.367$.
$-1/e$ is an absolute minimum value because:
(i) $f'(x) < 0$ for all $x < -1/2$ and $f'(x) > 0$ for all $x > -1/2$,
or
(ii) $f'(x)$ changes from negative to positive at $x = -1/2$, which is the only critical number.

\(\textbf{2(c)}\)
The range of $f$ is $[-1/e, \infty)$ or $[-0.367, \infty)$ or $[-0.368, \infty)$.

\(\textbf{2(d)}\)
$y’ = be^{bx} + b^2xe^{bx} = be^{bx}(1 + bx) = 0$
If $x = -1/b$, then $y = -1/e$ at $x = -1/b$.
$y$ has an absolute minimum value of $-1/e$ for all nonzero $b$.

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