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

The function 𝑔 is defined for π‘₯ > 0 with 𝑔(πœ‹) = 7, and 𝑔′(π‘₯) = 3π‘₯ +(sin 2/π‘₯). Use the line tangent to 𝑔 at π‘₯ = πœ‹ to approximate the value of 𝑔(3.1) and decide whether it over- or underestimates 𝑔(3.1).

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Video Transcript

The function 𝑔 is defined for π‘₯ is greater than zero with 𝑔 of πœ‹ equals seven, and 𝑔 prime of π‘₯ equals three π‘₯ plus sign two over π‘₯. Use the line tangent to 𝑔 at π‘₯ equals πœ‹ to approximate the value of 𝑔 of 3.1 and decide whether it over- or underestimates 𝑔 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‘Ž. The formula here comes directly from the equation for a straight line. Remember a straight line that passes through π‘₯ one, 𝑦 one and has a gradient of π‘š has the equation 𝑦 minus 𝑦 one equals π‘š times π‘₯ minus π‘₯ one. For some function 𝑓 of π‘₯ which has a point at π‘₯ equals π‘Ž, the 𝑦-coordinate will be given by 𝑓 of π‘Ž. And the gradient of the tangent line at this point will be given by 𝑓 prime of π‘Ž.

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

Now, our function is 𝑔. So I’ve rewritten this to say 𝑔 of π‘₯ is approximately equal to 𝑔 of π‘Ž plus 𝑔 prime of π‘Ž times π‘₯ minus π‘Ž. We’re looking to evaluate 𝑔 of 3.1. So π‘₯ is 3.1. We’re using the line tangent to 𝑔 at the point, where π‘₯ equals πœ‹. So π‘Ž is equal to πœ‹. And we can now substitute what we know into this formula. We can say that 𝑔 of 3.1 is approximately equal to 𝑔 of πœ‹ plus 𝑔 prime of πœ‹ times 3.1 minus πœ‹.

We’ve been told though that 𝑔 of πœ‹ is equal to seven. And we can work out 𝑔 prime of πœ‹ by using the formula for the derivative of 𝑔 of π‘₯. It’s three π‘₯ plus sin of two over π‘₯ which in turn gives us three times πœ‹ plus sin of two over πœ‹. So 𝑔 of 3.1 is approximately equal to seven plus three πœ‹ plus sin of two over πœ‹ times 3.1 minus πœ‹. Typing this into our calculator and we see that 𝑔 of 3.1 is approximately equal to 6.5832 and so on. Let’s say correct to three significant figures that 𝑔 of 3.1 is approximately equal to 6.58.

We’re not quite finished there. We need to decide whether this is an over- or underestimate of the actual value of 𝑔 of 3.1. Here, we use the fact that if our function is concave up in some interval around π‘₯ equals π‘Ž, then the approximation gives us an underestimate. And if the function is concave down in this interval, then it’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 𝑔. To do this, we’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, it’s a function of a function. It’s a composite function. So we’re going to use the chain rule. We’re going to let 𝑒 be equal to two over π‘₯ and let 𝑦 be equal to sin of 𝑒. In fact, let’s write 𝑒 as two times π‘₯ to the power of negative one.

The chain rule says that the derivative of 𝑦 with respect to π‘₯ is equal to d𝑦 by d𝑒 times d𝑒 by dπ‘₯. d𝑒 by dπ‘₯ is equal to negative two times π‘₯ to the negative two. And d𝑦 by d𝑒 is equal to cosine of 𝑒. So d𝑦 by dπ‘₯ is equal to negative two π‘₯ to the negative two times cosine of 𝑒. And of course, we let 𝑒 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 𝑔 prime prime of π‘₯.

We said that π‘Ž was equal to πœ‹. So we’re going to evaluate the second derivative at π‘₯ equals πœ‹. This gives us 2.83 which is greater than zero. And our function 𝑔 is concave up around the point where π‘₯ is equal to πœ‹, meaning our approximation must be an underestimate.

So 𝑔 of 3.1 is approximately equal to 6.58 correct to three significant figures and this is an underestimate.

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