Multiple choice calculus question

Question

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false.

If the graph of f is concave upward on (a, c) and concave downward on (c, b), where a < c < b, then f has an inflection point at x = c.

A)True. A point on the graph where the concavity changes is called an inflection point.

B) True. Let f(x) = −x³.

 Then f is concave upward on (−∞, 0)

 and concave downward on (0, ∞),

and has an inflection point at 0.    C) False. Let f(x) =  − 1

x

.

 Then f is concave upward on (−∞, 0)

 and concave downward on (0, ∞),

but f does not have an inflection point at 0. D)False. A point on the graph where the concavity changes is called a relative extremum. E) False. The point

c, f(c)

 must be either an inflection point or a relative extremum.

Answer 1

Well, to begin with, the question ought to specify that the function is continuous.

For instance, f(x) = tanx is concave up on (0 , π/2) and concave down on (π/2 , π) but it does not have an inflection point at x = π/2 because it is undefined there. This is very similar to a function's y-values changing signs without that function ever equalling 0 (in fact, that is exactly what y = tanx does at x = π/2 also).

Anyway, if we assume the function is continuous at x = c, then the point (c , f(c)) is indeed a point of inflection. That is the name for a point at which a function's graph changes concavity. If the second derivative is defined there, then f^"(c) = 0 and the second derivative will change signs there (since the sign of f^"(x) tells us the concavity of f). However, a function like y = ³√x changes concavity at the origin but neither f^'(0) nor f^"(0) are defined since (0,0) is a point of vertical tangency.

If the function f(x) is continuous at x = c, then (c , f(c)) is NOT ever a relative extremum.

Lastly, it is useful to note that f^'(c) need NOT = 0 for a point of inflection to occur. For instance, the function g(x) = x³ + x has a point of inflection at the origin but it is not a flat point of inflection, ie not a stationary point like it is for y = x³. Instead, g^'(x) = 3x² + 1 and g^'(0) = 1. So the origin represents the point at which g(x) is increasing at the slowest rate it ever does (namely 1). Because it goes from increasing at a slower and slower rate to increasing at a faster and faster rate at x = 0, g(0) is a point of inflection on the graph of g(x).