Question

if the figure below is the graph of the derivative f, answer the following:
where do the points of inflection of f occur?
on which interval(s) is f concave down?

Explanation:

Step1: Recall the relationship between \(f\) and \(f'\)

A point of inflection of \(y = f(x)\) occurs where the concavity of \(f\) changes. This is equivalent to where the derivative \(f'\) changes its concavity. The concavity of \(y = f(x)\) is related to the second - derivative \(f''\). But since we have the graph of \(f'\), the concavity of \(f\) is determined by the slope of \(f'\). A point of inflection of \(f\) occurs where the slope of \(f'\) (i.e., \(f''\)) changes sign. Looking at the graph of \(f'\), the slope of \(f'\) changes sign at \(x = 1\) and \(x=6\). So the points of inflection of \(f\) occur at \(x = 1\) and \(x = 6\).

Step2: Determine the interval where \(f\) is concave down

The function \(y = f(x)\) is concave down when \(f''(x)<0\). Since \(f''(x)\) is the slope of \(y = f'(x)\), \(f(x)\) is concave down when the slope of \(f'(x)\) is negative. Looking at the graph of \(f'\), the slope of \(f'(x)\) is negative on the interval \((1,6)\)

Answer:

• Points of inflection of \(f\) occur at \(x = 1\) and \(x = 6\)
• \(f\) is concave down on the interval \((1,6)\)