AP Calculus AB/BC FRQ Practice

♾️AP Calculus AB/BC

Practice FRQ 1 of 81/8

1. The following functions are defined for this question: $f(x) = x + 2$

$G(x) = \begin{cases} \frac{x^2-4}{x-2} & x ≠ 2 \\ 4 & x = 2 \end{cases}$

$k(x) = 4$

Water flows through a filter system. For $x ≥ 0$ , the function $F$ defined by $F(x)=\frac{x^2-4}{x-2}$ for $x ≠ 2$ models the flow rate, in liters per minute, when the pressure setting is $x$ . The graph of $F$ for $0 ≤ x ≤ 6$ is shown in Figure 1. The point at $x=2$ is not included on the graph because $F(2)$ is undefined.

$f(x) = x + 2$
$G(x) = \begin{cases} \frac{x^2-4}{x-2} & x ≠ 2 \\ 4 & x = 2 \end{cases}$
$k(x) = 4$

Figure 1. Graph of F on 0 ≤ x ≤ 6, showing the removable discontinuity at x = 2 (F(2) undefined).

Single clean 2D Cartesian graph on a white background.

Axes (must be exact):
- Horizontal axis labeled "Pressure setting, x" with x-values from 0 to 6 inclusive. Tick marks at every 1 unit, with numeric labels shown at 0, 1, 2, 3, 4, 5, 6. The origin is labeled "0" at the axes intersection.
- Vertical axis labeled "Flow rate, F(x) (liters/min)" with y-values from 0 to 8 inclusive. Tick marks at every 1 unit, with numeric labels shown at 0 through 8.
- Put arrows on the positive ends of both axes.
- No gridlines.

Curve / function shown (must match the algebraic simplification):
- Plot the graph of the function F(x) = (x^2 − 4)/(x − 2) on the interval 0 ≤ x ≤ 6, but DO NOT include the point at x = 2.
- The plotted curve must be a perfectly straight solid line (not curved) everywhere it is drawn, with constant positive slope.
- The line must pass through the y-axis at y = 2 when x = 0 (so the left boundary of the plotted line touches the vertical axis at the tick labeled 2).
- The same straight line must also pass through y = 8 when x = 6 (so the right boundary of the plotted line reaches the tick labeled 8 exactly at x = 6).

Discontinuity / hole (critical feature):
- At the vertical tick mark labeled x = 2, remove the point from the line and show an open circle (a hollow marker) exactly on the line at the y-value labeled 4.
- There must be no filled point anywhere above or below this open circle at x = 2 (so the graph clearly indicates the function is undefined there).

Segment behavior (explicit):
- Left segment: from x = 0 up to but not including x = 2, draw the straight solid line increasing steadily; it ends visually at the open circle.
- Right segment: from just greater than x = 2 through x = 6, continue the same straight solid line with identical slope; it begins visually immediately after the open circle and continues to the right boundary.

Styling:
- Use a solid black line of medium thickness for the graph.
- Use a black open circle marker at the hole; the open circle should be visibly larger than the line thickness so the missing point is unmistakable.
- No legend and no title text inside the plot area (caption is outside the plot).

Use the graph of $F$ to estimate

$\displaystyle \lim_{x\to 2}F(x).$

Show the work that leads to your answer.

Find

$\displaystyle \lim_{x\to 2} \frac{x^2-4}{x-2}$

using limit theorems and/or equivalent expressions. Show the algebraic work that leads to your answer, and write your final answer using correct limit notation.

Write a limit expression that represents the instantaneous rate of change of $F$ at $x=3$ in terms of average rates of change over intervals containing $x=3$ . Then use your calculator to approximate this instantaneous rate of change.

A new function $G$ is defined by

$G(x)=\begin{cases}\frac{x^2-4}{x-2}, & x≠ 2\\ k, & x=2\end{cases}$

where $k$ is a constant. Find the value of $k$ that makes $G$ continuous at $x=2$ . Justify your answer using the definition of continuity at a point. The function $G$ matches $F$ for all $x≠ 2$ but assigns the value $k$ at $x=2$ . Continuity at $x=2$ requires that $\lim_{x\to 2}G(x)=G(2)$ .