# Video: AP Calculus AB Exam 1 β’ Section I β’ Part B β’ Question 85

The function π is defined for π₯ > 0 with π(π) = 7, and πβ²(π₯) = 3π₯ +(sin 2/π₯). Use the line tangent to π at π₯ = π to approximate the value of π(3.1) and decide whether it over- or underestimates π(3.1).

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### Video Transcript

The function π is defined for π₯ is greater than zero with π of π equals seven, and π prime of π₯ equals three π₯ plus sign two over π₯. Use the line tangent to π at π₯ equals π to approximate the value of π of 3.1 and decide whether it over- or underestimates π of 3.1.

Linear approximation or tangent line approximation is the idea of using the equation of the tangent line to approximate values of a function for π₯ near some point π₯ equals π. The formula here comes directly from the equation for a straight line. Remember a straight line that passes through π₯ one, π¦ one and has a gradient of π has the equation π¦ minus π¦ one equals π times π₯ minus π₯ one. For some function π of π₯ which has a point at π₯ equals π, the π¦-coordinate will be given by π of π. And the gradient of the tangent line at this point will be given by π prime of π.

We can replace these values in our equation for a straight line. And we see that we have π¦ minus π of π equals π prime of π times π₯ minus π. And adding π of π to both sides of our equation, we get π¦ equals π of π plus π prime of π times π₯ minus π. We can use this. And we can say that for π₯ near π of π₯ is approximately equal to π of π plus π prime of π times π₯ minus π.

Now, our function is π. So Iβve rewritten this to say π of π₯ is approximately equal to π of π plus π prime of π times π₯ minus π. Weβre looking to evaluate π of 3.1. So π₯ is 3.1. Weβre using the line tangent to π at the point, where π₯ equals π. So π is equal to π. And we can now substitute what we know into this formula. We can say that π of 3.1 is approximately equal to π of π plus π prime of π times 3.1 minus π.

Weβve been told though that π of π is equal to seven. And we can work out π prime of π by using the formula for the derivative of π of π₯. Itβs three π₯ plus sin of two over π₯ which in turn gives us three times π plus sin of two over π. So π of 3.1 is approximately equal to seven plus three π plus sin of two over π times 3.1 minus π. Typing this into our calculator and we see that π of 3.1 is approximately equal to 6.5832 and so on. Letβs say correct to three significant figures that π of 3.1 is approximately equal to 6.58.

Weβre not quite finished there. We need to decide whether this is an over- or underestimate of the actual value of π of 3.1. Here, we use the fact that if our function is concave up in some interval around π₯ equals π, then the approximation gives us an underestimate. And if the function is concave down in this interval, then itβs an overestimate. And we can determine its concavity by finding the second derivative of our function at this point.

So we need to find the second derivative of our function π. To do this, weβre going to differentiate the function three π₯ plus sin of two over π₯. And the derivative of three π₯ with respect to π₯ is simply three. But what about the derivative of sin of two over π₯ with respect to π₯? Well, itβs a function of a function. Itβs a composite function. So weβre going to use the chain rule. Weβre going to let π’ be equal to two over π₯ and let π¦ be equal to sin of π’. In fact, letβs write π’ as two times π₯ to the power of negative one.

The chain rule says that the derivative of π¦ with respect to π₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. dπ’ by dπ₯ is equal to negative two times π₯ to the negative two. And dπ¦ by dπ’ is equal to cosine of π’. So dπ¦ by dπ₯ is equal to negative two π₯ to the negative two times cosine of π’. And of course, we let π’ be equal to two π₯ to the power of negative one. And so, we now have the derivative of sin of two over π₯. And in turn, we have the second derivative π prime prime of π₯.

We said that π was equal to π. So weβre going to evaluate the second derivative at π₯ equals π. This gives us 2.83 which is greater than zero. And our function π is concave up around the point where π₯ is equal to π, meaning our approximation must be an underestimate.

So π of 3.1 is approximately equal to 6.58 correct to three significant figures and this is an underestimate.