An inflection point occurs where the second derivative is zero or undefined and a sign change occurs in which of the following?

An inflection point occurs where the second derivative is zero or undefined and a sign change occurs in which of the following?

• A. f''(c) > 0
• B. f''(c) = 0 OR f''(x) is undefined AND f''(c) switches signs
• C. f'''(c) = 0
• D. f''(c) < 0

Answer: (B)

Inflection points are where the graph changes concavity, which means the second derivative must switch sign as you move through that x-value. The second derivative being zero or undefined at the point provides a chance for a concavity change, but the actual inflection occurs only if the sign of f'' changes across the point. So the condition that captures both ideas is: f'' is zero or undefined at the point and, as you pass through that point, f'' changes sign. That’s why a case like f''(c) > 0 or f''(c) < 0 isn’t enough—those indicate a fixed concavity on one side, not a switch. And a statement about the third derivative being zero isn’t tied to concavity changes. A concrete example helps: f(x) = x^3 has an inflection at 0 because f''(x) = 6x changes from negative to positive as x passes through 0, and f''(0) = 0. But f(x) = x^4 has f''(0) = 0 yet f''(x) = 12x^2 stays nonnegative on both sides, so there is no sign change and no inflection.

Inflection points are where the graph changes concavity, which means the second derivative must switch sign as you move through that x-value. The second derivative being zero or undefined at the point provides a chance for a concavity change, but the actual inflection occurs only if the sign of f'' changes across the point. So the condition that captures both ideas is: f'' is zero or undefined at the point and, as you pass through that point, f'' changes sign.

That’s why a case like f''(c) > 0 or f''(c) < 0 isn’t enough—those indicate a fixed concavity on one side, not a switch. And a statement about the third derivative being zero isn’t tied to concavity changes. A concrete example helps: f(x) = x^3 has an inflection at 0 because f''(x) = 6x changes from negative to positive as x passes through 0, and f''(0) = 0. But f(x) = x^4 has f''(0) = 0 yet f''(x) = 12x^2 stays nonnegative on both sides, so there is no sign change and no inflection.