AP Calculus BC 5.9 Connecting a Function, Its First Derivative, and Its Second Derivative - FRQs - Exam Style Questions
Calc-Ok Question
Let f be the continuous function defined on [-4, 3] whose graph, consisting of three line segments and a semicircle centered at the origin, is given above. Let g be the function given by \(g(x)=\int_1^x f(t) dt\).
(a) Find the values of g(2) and g(-2).
(b) For each of g'(-3) and g”(-3) , find the value or state that it does not exist.
(c) Find the x-coordinate of each point at which the graph of g has a horizontal tangent line. For each of these points, determine whether g has a relative minimum, relative maximum, or neither a minimum nor a maximum at the point. Justify your answers.
(d) For -4 < x < 3, find all values of x for which the graph of g has a point of inflection. Explain your reasoning.
Most-appropriate topic codes (CED):
• TOPIC 5.9: Connecting a Function, Its First Derivative, and Its Second Derivative — part (b)
• TOPIC 5.4: Using the First Derivative Test to Determine Relative (Local) Extrema — part (c)
• TOPIC 5.2: Extreme Value Theorem, Global Versus Local Extrema, and Critical Point — part (d)
▶️ Answer/Explanation
(a)
By definition \(g(x)=\int_{1}^{x} f(t)\,dt\). Thus \(g(2)=\int_{1}^{2} f(t)\,dt\) and \(g(-2)=\int_{1}^{-2} f(t)\,dt=-\int_{-2}^{1} f(t)\,dt\).
From the geometry of the graph on \([1,2]\), the region is a right triangle below the \(t\)-axis with base \(1\) and height \(\tfrac12\), so the signed area is \(-\tfrac12\cdot 1\cdot \tfrac12=-\tfrac14\). Hence \(\boxed{g(2)=-\tfrac14}\).
On \([-2,1]\), the net signed area breaks into (i) a triangular piece totaling \(-\tfrac{3}{2}\) and (ii) a semicircle of radius \(1\) below the axis with area \(-\tfrac{\pi}{2}\). Therefore \(\int_{-2}^{1} f(t)\,dt= -\tfrac{3}{2}-\tfrac{\pi}{2}\), and so
\(g(-2)=-\!\left(-\tfrac{3}{2}-\tfrac{\pi}{2}\right)=\boxed{\tfrac{\pi}{2}-\tfrac{3}{2}}.\)
(b)
Since \(g'(x)=f(x)\) (FTC Part 1), we have \(g'(-3)=f(-3)\). From the graph, \(f(-3)=2\) ⇒ \(\boxed{g'(-3)=2}\).
Also \(g”(x)=f'(x)\) wherever \(f\) is differentiable. At \(x=-3\) the graph of \(f\) is linear with slope \(1\), so \(f'(-3)=1\) ⇒ \(\boxed{g”(-3)=1}\).
(c)
Horizontal tangents of \(g\) occur where \(g'(x)=0\), i.e., where \(f(x)=0\). From the graph this happens at \(x=-1\) and \(x=1\).
Classify using sign changes of \(g'(x)=f(x)\):
• At \(x=-1\), \(f\) changes from \(+\) to \(-\) (so \(g’\) goes \(+\to-\)), giving a relative maximum of \(g\) at \(x=-1\).
• At \(x=1\), \(f\) touches/crossing behavior yields no sign change of \(f\) across \(x=1\); hence \(g’\) does not change sign ⇒ neither a max nor a min at \(x=1\).
Conclusion: \(\boxed{x=-1\ \text{(relative max)},\ x=1\ \text{(neither)}}\).
(d)
Points of inflection of \(g\) occur where \(g”(x)=f'(x)\) changes sign (i.e., where the slope of \(f\) changes from increasing to decreasing or vice versa).
From the piecewise-linear/semicircular graph, this happens at the transition/turning points \(x=-2,\ 0,\ 1\). At each of these, the slope of \(f\) changes sign, so \(f'(x)\) changes sign and \(g”\) changes sign as well.
Therefore the graph of \(g\) has points of inflection at \(\boxed{x=-2,\ 0,\ 1}\).