Full response questions

Question

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The figure above shows the graph of f ', the derivative of a twice-differentiable function f, on the interval[–3, 4]. Answer the following questions based on the graph of f ′.

a) Find all x-coordinates at which f has critical values and identify them as a relative minimum,

relative maximum, or neither. Justify your answer.

b) On what open intervals contained in [–3, 4] is the graph of f both concave up and decreasing?

Justify your answer.

c) Find the x-coordinates of all points of inflection on the graph of f

Answer 1

To answer this question, the trick is to almost ignore the graph you’re given. You only need a few point of it anyway. Instead, imagine the graph of a function, any function, that has maxima and minima and points of inflection (these are what ‘critical points’ means). A maximum is at the top of a bump, a minimum at the bottom, and a point of inflection is when the graph changes its mind and decides to not go up (or down) after all.

Now, at these points, the slope of your graph is flat, or zero. Think about it. At the very top, it flattens before falling. Same thing at the other two. This means that the derivative is zero, since that’s what a derivative is: the value of the slope of the graph of a function. So what you want to find is when the derivative is zero.

This is when you do look at the graph you’re given. Can you see the three points when it goes through or touches zero? At x= -2, 1, and 3!

The problem also asks you to decide what type of critical point you have. Again, ignore that graph and think instead of your imaginary function. When you have a max, you climb up to it, then you fall. That is, the slope is first positive, then negative. So you need to see whether at any of your x values, the graph of your derivative is first positive, then negative. If you find one such point, that will be a max in the graph.

Similarly, you will have a min if your derivative is negative, then positive. Can you see that around one of your x values in the graph given?

Finally, a point of inflection will continue being either positive or negative, since that’s when the graph changed its mind about switching. There’s one x value where your derivative graph continues to be negative after touching zero. Can you see it?

As to the question about concavity, you need to again go back to your imaginary function graph. Since it says it’s decreasing, that means it’s derivative will be negative, because the graph will be sloping downwards. But the concavity being up means that it’s the downward slope of a smiling face, approaching a min (or point of inflection), not a max. So, going back to your derivative graph, we want to find the parts in the negative part of the axes, but only those approaching the min or the point of inflection, at x=1 and x=-2 respectively. In particular, you want to look for parts of the graph that have a negative slope, because that means that the slope of the original graph is decreasing. Thus you can see that only (-3,-2) and (0,1) work