Home / AP Calculus AB: 5.6 Determining Concavity of Functions over Their  Domains – Exam Style questions with Answer- MCQ

AP Calculus AB: 5.6 Determining Concavity of Functions over Their  Domains – Exam Style questions with Answer- MCQ

Question

 If \(x+3y^{\frac{1}{3}}=y\) ,what is \(\frac{\mathrm{d} y}{\mathrm{d} x}\) at the point (2,8) ?

A \(\frac{1}{3}\)

B \(\frac{3}{4}\)

C \(\frac{5}{4}\)

D \(\frac{4}{3}\)

▶️Answer/Explanation

Ans:D

The chain rule is the basis for implicit differentiation.\(1+y^{-\frac{2}{3}}\frac{\mathrm{d} y}{\mathrm{d} x}\)

The point (2,8)  is on the curve since x=2 and y=8 satisfy the equation. At this point,

\(1+\frac{1}{4}\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{\mathrm{d} y}{\mathrm{d} x}\Rightarrow \frac{3}{4}\frac{\mathrm{d} y}{\mathrm{d} x}=1\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{4}{3}\)

Question

Let f be the function defined by \(f(x)=-3+6x^{2}-2x^{3}\) What is the largest open interval on which the graph of f is both concave up and increasing?

A (0, 1)

B (1, 2)

C (0, 2)

D (2,)

▶️Answer/Explanation

Ans:A

Question

The twice-differentiable functions f, g, and h have second derivatives given above. Which of the functions f, g, and h have a graph with exactly two points of inflection?

A g only

B h only

C f and g only

D f, g, and h

▶️Answer/Explanation

Ans:C

Question

Let g be the function given by  g(x)=’\(\int ^{x}_{3}(t^{2}-5t-14)dt\) . What is the x-coordinate of the point of inflection of the graph of g?
A -2
B\(\frac{5}{2}\)
C3
D7

▶️Answer/Explanation

Ans:B
To find the point of inflection of the graph of g,determine where g″ changes sign.
g′(x)\(=x^{2}-5x-14\)
g′′(x)=2x−5
Then g′′(x)=0 at \(x=\frac{5}{2}\). Since g′′(x)<0 for \(x<\frac{5}{2}\) and g′′(x)>0 for \(x>\frac{5}{2}\), the graph of g changes concavity at \(x=\frac{5}{2}\) and therefore, the graph of g has a point of inflection at \(x=\frac{5}{2}\).

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