AP Calculus BC 1.1 Introducing Calculus: Can Change Occur at an Instant? Exam Style Questions - FRQ
Question
(a) Topic-5.4-Using the First Derivative Test to Determine Relative Local Extrema
(b) Topic-5.4-Using the First Derivative Test to Determine Relative Local Extrema
(c) Topic-5.5-Using the Candidates Test to Determine Absolute Global Extrema
(d) Topic-1.1-Introducing Calculus Can Change Occur at an Instant
The figure above shows the graph of the piecewise-linear function f. For -4 ≤ x ≤ 12, the function g is defined by \(g(x)=\int_{2}^{x}f(t)dt.\)
(a) Does g have a relative minimum, a relative maximum, or neither at x = 10 ? Justify your answer.
(b) Does the graph of g have a point of inflection at x = 4 ? Justify your answer.
(c) Find the absolute minimum value and the absolute maximum value of g on the interval -4 ≤ x ≤ 12. Justify your answers.
(d) For -4 ≤ x ≤ 12 , find all intervals for which g (x) ≤ 0.
▶️Answer/Explanation
\(\textbf{3(a)} \text{The function } g \text{ has neither a relative minimum nor a relative maximum at } x = 10 \text{ since } g'(x) = f(x) \text{ and } f(x) \leq 0 \text{ for } 8 \leq x \leq 12.\)
\(\textbf{3(b)} \text{The graph of } g \text{ has a point of inflection at } x = 4 \text{ since } g'(x) = f(x) \text{ is increasing for } 2 \leq x \leq 4 \text{ and decreasing for } 4 \leq x \leq 8.\)
\(\textbf{3(c) } g'(x) = f(x) \text{ changes sign only at } x = -2 \text{ and } x = 6.\)
\(
\begin{array}{|c|c|}
\hline
x & g(x) \\
\hline
-4 & -4 \\
-2 & -8 \\
6 & 8 \\
12 & -4 \\
\hline
\end{array}
\)
\(\text{On the interval } -4 \leq x \leq 12, \text{ the absolute minimum value is } g(-2) = -8 \text{ and the absolute maximum value is } g(6) = 8.\)
\(\textbf{3(d) } g(x) \leq 0 \text{ for } -4 \leq x \leq 2 \text{ and } 10 \leq x \leq 12.\)