When searching for inflection points on a function, you can narrow your search by identifying numbers where the function is continuous (from both directions) and the second derivative is either zero or undefined. (By definition an inflection point cannot occur at a number where the function is not continuous from both directions.) You can then build a sign table for the second derivative that implies the concavity of the given function.
When performing this analysis, you need to simplify the second derivative formula in the same way you simplify the first derivative formula when looking for critical numbers and local extreme points.
Yolanda was given this information and asked to find the inflection points on \(y\text{.}\) The first thing Yolanda wrote was, βThe critical numbers of \(y\) are \(\sqrt{3}\) and \(-\sqrt{3}\text{.}\)β Explain to Yolanda why this is not true.
The function \(y\) has a vertical asymptote at \(-3\text{.}\) Given that fact, it was impossible that \(y\) would have an inflection point at \(-3\text{.}\) Why, then, did we nevertheless break the interval \(\ointerval{-\infty}{-\sqrt{3}}\) at \(-3\) when creating our concavity table?
Find, and completely simplify, the formula for the second derivative of the function. It is not necessary to simplify the formula for the first derivative of the function.
The second derivative of the function \(\fe{w}{t} = t^{1.5} - 9t^{0.5}\) is \(\fe{\sd{w}}{t} = \frac{3(t+3)}{4t^{1.5}}\) yet \(w\) has no inflection points. Why is that?