Video: APCALC02AB-P1B-Q38-980173932895

The graph of 𝑓′, the derivative of the function 𝑓, is shown. Which of the following statements must be true? [A] 𝑓 has a relative maximum at π‘₯ = βˆ’1. [B] The graph of 𝑓 has a point of inflection at π‘₯ = βˆ’2. [C] The graph of 𝑓 is concave down for 0 < π‘₯ < 4.

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Video Transcript

The graph of 𝑓 prime, the derivative of the function 𝑓, is shown. Which of the following statements must be true? I) 𝑓 has a relative maximum at π‘₯ equals negative one. II) The graph of 𝑓 has a point of inflection at π‘₯ equals negative two. III) The graph of 𝑓 is concave down for zero is less than π‘₯ is less than four.

Let’s take each of these statements in turn. In statement one, we’re concerned about a relative maximum. Now, a relative maximum is an example of a critical point. And we know that the critical points of a function occur when its first derivative, 𝑓 prime of π‘₯, is equal to zero or is undefined. From the graph, we see that 𝑓 prime of negative one is indeed equal to zero, because the graph of 𝑓 prime crosses the π‘₯-axis at this point. And therefore, the function does have some kind of critical point at π‘₯ equals negative one.

In order to determine what type of critical point our function has, we need to consider the sign of the slope of the function either side of the critical point. We see that the graph of the first derivative is above the π‘₯-axis for π‘₯-values to the left of negative one. And it’s below the π‘₯-axis for π‘₯-values to the right of negative one. The slope therefore changes from positive to negative, which confirms that this critical point is indeed a relative maximum. Statement I) is therefore true. Let’s consider statement II), which concerns points of inflection.

A point of inflection is a point on a curve where its concavity changes, either from concave upwards to concave downwards or vice versa. It’s also the case that, at points of inflection, the second derivative of the function 𝑓 double prime of π‘₯ will be equal to zero. So in order to determine whether our function has a point of inflection at π‘₯ equals negative two, we need to consider 𝑓 double prime of negative two. We need to recall that the second derivative gives the slope of the first derivative. So by sketching in a tangent to the graph of 𝑓 prime at π‘₯ equals negative two, we can consider its slope. And hence the value of the second derivative. We see that the tangent is not horizontal. And it has a negative slope. And therefore, it isn’t the case that 𝑓 double prime of negative two is equal to zero. The second statement is therefore false although it does look like the graph of 𝑓 prime could have a point of inflection at π‘₯ equals negative two, but not the graph of 𝑓 itself.

Finally, let’s consider statement III). This statement concerns the concavity of the curve and in particular whether the graph of 𝑓 is concave down in a particular interval. We recall that a graph is concave down in a particular interval if the tangents to the curve lie above the curve itself in that interval. We also recall that when a graph is concave down, the slope of its tangent is decreasing. And therefore, the value of the first derivative 𝑓 prime of π‘₯ will also be decreasing. From the graph of 𝑓 prime, we see that whilst this might be true for π‘₯-values between zero and one, it isn’t true for π‘₯-values between one and four, because in this interval, the graph of 𝑓 prime is climbing, which means that the value of 𝑓 prime of π‘₯ is increasing. This tells us then that the graph of 𝑓 is not concave down on the whole interval zero is less than π‘₯ is less than four. And so statement III) is false.

We could answer the question then by concluding that the only one of the three statements which must be true is statement I).

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