Video: Comparing the Values of a Function, and Its First and Second Derivatives for a Particular π‘₯ Value When the Function Is Defined as a Definite Integral of Another Function

The graph of a differentiable function 𝑓 is shown. If 𝑔(π‘₯) = ∫_(0)^(π‘₯) 𝑓(𝑑) d𝑑, which of the following is true? a) 𝑔′(2) > 𝑔(2) > 𝑔″(2) b) 𝑔′(2) > 𝑔″(2) > 𝑔(2) c) 𝑔″(2) > 𝑔′(2) > 𝑔(2) d) 𝑔″(2) > 𝑔(2) > 𝑔′(2)

03:50

Video Transcript

The graph of a differentiable function 𝑓 is shown. If 𝑔 of π‘₯ is equal to the integral from zero to π‘₯ of 𝑓 of 𝑑 d𝑑, which of the following is true? a) 𝑔 prime of two is greater than 𝑔 of two is greater than 𝑔 double prime of two. Where 𝑔 prime of two is d𝑔 by dπ‘₯ at π‘₯ equals two. 𝑔 of two is 𝑔 at π‘₯ equals two. And 𝑔 double prime of two is the second derivative of 𝑔 with respect to π‘₯ at π‘₯ equals two. b) 𝑔 prime of two is greater than 𝑔 double prime of two is greater than 𝑔 of two. c) 𝑔 double prime of two is greater than 𝑔 prime of two is greater than 𝑔 of two. Or d) 𝑔 double prime of two is greater than 𝑔 of two is greater than 𝑔 prime of two.

The first thing to notice is that all four statements are inequalities containing the same three expressions. 𝑔 of two, 𝑔 prime of two, and 𝑔 double prime of two. To decide which of these statements is true, we don’t necessarily need to find the values of 𝑔, 𝑔 prime, and 𝑔 double prime at π‘₯ equals two. We simply need to compare their magnitudes.

Let’s look first at the function 𝑔 at π‘₯ equals two. If we substitute π‘₯ equals two into the given integral, we find that 𝑔 of two is the integral between zero and two of 𝑓 of 𝑑 d𝑑. Remember that the graph we’ve been given is the graph of the function 𝑓. And we’re looking at a definite integral of 𝑓 between π‘₯ is zero and π‘₯ is two. The value of the integral is actually the area between these limits. And since in our case the area is below the horizontal axis, the result of the integral will be negative. We can therefore say that 𝑔 of two is negative. That is, 𝑔 of two is less than zero.

Now we know something about 𝑔 of two. Let’s look at 𝑔 prime of two. For 𝑔 prime of two, we can use the fundamental theorem of calculus. This tells us that if a function 𝑓 is continuous on the closed interval π‘Žπ‘, where π‘₯ lies between π‘Ž and 𝑏. If capital 𝐹 of π‘₯ is equal to the integral from π‘Ž to π‘₯ of 𝑓 of 𝑑 d𝑑. Then capital 𝐹 prime of π‘₯ is equal to 𝑓 of π‘₯. In our case, π‘Ž is equal to zero. And capital 𝐹 of π‘₯ is equal to 𝑔 of π‘₯. So by the fundamental theorem of calculus, 𝑔 prime of π‘₯ is 𝑓 of π‘₯. If we now substitute π‘₯ equal to two, we have 𝑔 prime of two is equal to 𝑓 of two. If we look for 𝑓 of two on our graph, we can see that 𝑓 of two is equal to zero. Because 𝑓 crosses the π‘₯-axis at π‘₯ equals two. So in fact 𝑔 prime of two is equal to zero.

So far so good. We have 𝑔 of two less than zero. And 𝑔 prime of two equal to zero. So now let’s look at 𝑔 double prime of two. We know that 𝑔 prime of π‘₯ is equal to 𝑓 of π‘₯. And if we differentiate this with respect to π‘₯, we have d𝑔 prime by dπ‘₯ is equal to 𝑔 double prime of π‘₯ is equal to 𝑓 prime of π‘₯. Now, 𝑓 prime of π‘₯ is the slope of the function 𝑓. So what does this look like at π‘₯ equals two? Let’s look on our graph for the slope of the function 𝑓 at π‘₯ equals two. The slope of the tangent to 𝑓 at π‘₯ equals two is positive. This means that 𝑔 double prime of two, which is 𝑓 prime of two, must be positive also. That is, 𝑔 double prime of two is greater than zero.

We have all the information we need now to decide which of the four statements is true. We know that 𝑔 double prime of two is greater than zero. 𝑔 prime of two is equal to zero. And 𝑔 of two is less than zero. So 𝑔 double prime of two is greater than 𝑔 prime of two. And 𝑔 prime of two is greater than 𝑔 of two. And this corresponds to our statement c.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.