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

Question

The first derivative of the function f is defined by \({f}'(x) = (x^2 +1)sin(3x – 1)\) for \(-1.5 < x < 1.5\). On which of the following intervals is the graph of f concave up?

A \((−1.5, −1.341)\) and \((−0.240, 0.964)\)

B \((−1.341, −0.240)\) and \((0.964, 1.5)\)

C \((−0.714, 0.333)\) and \((1.381, 1.5)\)

D \((−1.5, −0.714) and (0.333, 1.381)\)

Answer/Explanation

 

Question

The graph of y  = ( fx) on the closed interval [2,7] is shown above. How many points of  inflection does this graph have on this interval?
(A) One                          (B) Two                            (C) Three                      (D) Four                                 (E) Five

Answer/Explanation

Ans:C

 

Question

Let f be the function given by\(f(x)=cosx(2x)+ln(3x)\). What is the least value of x at which the   graph of f changes concavity?

(A) 0.56                                     (B) 0.93                                       (C) 1.18                                               (D) 2.38                                               (E) 2.44

Answer/Explanation

Ans:B

 

Question

 At what value of x does the function \(y=1.2x^{2}-e^{.4x}\)  change concavity?
(A) x = 2.5 ln 6
(B) x = 6 ln 2.4
(C) x = 2.4 ln 12
(D) x = 2.5 ln 15

Answer/Explanation

Ans:(D) 

\(y’=2.4x-.4e^{.4x}\)                                                                                                                                                                                                                                                                                                                                                             \( y”=2.4-.16e^{.4x}\) The function can only change concavity when y ″ = 0

\(0=2.4-.16e^{.4x}\)
\(.16e^{.4x}=2.4\)
\(e^{.4x}=\frac{2.4}{.16}\)
\(e^{.4x}=15\)
\(0.4x=\ln 15\)

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