<p>For the answer to 5b, I originally tried to find the second derivative of function g, which is f ' (t), and then set that equal to 0 to find the point of inflection. The graph of f though has no zero value for the slope of the graph of f. What is the wrong with my reasoning and can someone explain the given answer please?</p>
<p>I assume you meant 4b</p>
<p>There are 4 points on g where the function is not differentiable, so you have to check those in addition to the points where g’=0. Inflection happens when f goes from concave up to down or vice versa, or f’ (i.e., g) goes from increasing to decreasing, or vice versa</p>
<p>How is f’ equal to g? I thought that g’ was equal to f. My thinking is that i need to find g’’ and set that equal to zero to find the point of inflections but I’m wrong for some reason.</p>
<p>Whoops, you’re right. But my point about critical points still stands.</p>
<p>g’’=f’, which must equal 0 OR NOT EXIST for an IP.</p>
<p>Oh wait, so to find an inflection point, the second derivative can also equal undefined? I thought the second derivative can only equal to zero for an inflection point.</p>
<p>Just think of a point of inflection as a critical number. Just like with critical number’s f’, points of inflection occur when f’’=0 or DNE/undefined. Only difference is that for it to be a point of inflection it also has to change from positive to negative or vice versa.</p>
<p>With the scoring, too, mentioning the sign change for the second derivative is the only part that matters. Finding where the second derivative is either zero or undefined only tells you locations where inflection points MIGHT change signs, not where they necessarily do.</p>