AP Calculus AB Concavity and the Second Derivative Quiz: 10 Questions with Explanations

A concave upB concave downC decreasingD linear

A positive second derivative means the slope is increasing, so the graph is concave up.

A it has a relative maximumB it has an inflection pointC it has a relative minimumD it has a vertical asymptote

The concavity changes at x = 4 because f″ changes sign, so the graph has an inflection point.

A \(x=2\)B \(x=0\)C \(x=\sqrt{2}\)D \(x=-6\)

f″ = 6x, which changes sign at x = 0.

A a relative minimumB an inflection pointC no conclusionD a relative maximum

A negative second derivative at a critical number means concave down, hence a relative maximum.

A concave up everywhereB concave down everywhereC concave up for x > 2 onlyD it has an inflection point at x = 2

f″ = 2 > 0 for all x, so the graph is always concave up with no inflection points.

A conclude a minimumB use the First Derivative TestC conclude a maximumD conclude an inflection point

The Second Derivative Test is inconclusive when f″(c) = 0, so check the sign of f′ around c.

A it is concave downB it is decreasingC it is concave upD it has a maximum

f′ increasing means f″ > 0, which means f is concave up.

A \((0,0)\)B \((2,-4)\)C \((1,-2)\)D \((1,2)\)

f″ = 6x – 6 = 0 at x = 1 and f(1) = -2. (0, 0) and (2, -4) are the critical points.

A \(f(x)=x^{2}\)B \(f(x)=x^{3}\)C \(f(x)=\sin x\)D \(f(x)=x^{5}\)

f″ = 2 never changes sign. The others have f″ that changes sign (at 0, at multiples of pi, and at 0).

A it has a relative maximum at x = 3B it has a relative minimum at x = 3C it is undefined at x = 3D it has an inflection point at x = 3

A local extremum of f′ means f″ changes sign there, so f changes concavity at x = 3.