# Mistakes when finding inflection points: not checking candidates | AP Calculus AB | Khan Academy | Summary and Q&A

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September 7, 2017
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Mistakes when finding inflection points: not checking candidates | AP Calculus AB | Khan Academy

## TL;DR

Olga's solution for finding inflection points is incorrect because she incorrectly concludes that there is an inflection point at x equals two based on the fact that the second derivative is zero at that point.

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### Q: What did Olga do correctly in her solution for finding inflection points?

Olga correctly found the first and second derivatives of the function and correctly determined that x equals two is a critical point.

### Q: What is Olga's mistake in her approach to finding inflection points?

Olga's mistake is concluding that there is an inflection point at x equals two solely based on the fact that the second derivative is zero at that point. She fails to test for concavity change by analyzing the sign of the second derivative on both sides of x equals two.

### Q: How does one determine if there is an inflection point at a given point?

To determine if there is an inflection point at a given point, one must check if the sign of the second derivative changes as you move from values less than the given point to values greater than the given point. If the signs do not change, there is no inflection point.

### Q: Can inflection points occur when the second derivative is zero?

Inflection points can occur when the second derivative is zero, but the second derivative being zero at a point does not guarantee the presence of an inflection point. The change in concavity must be checked for to confirm the existence of an inflection point.

## Summary & Key Takeaways

• Olga attempts to find inflection points by finding the first and second derivatives of the given function.

• She correctly finds the first and second derivatives, and determines that x equals two is a critical point.

• However, she incorrectly concludes that there is an inflection point at x equals two based solely on the fact that the second derivative is zero at that point.