Video: Estimating the Value of a Function at a Point Using Tangent Line Approximation

For the function 𝑓, 𝑓′(π‘₯) = (1/2) π‘₯ + 3 and 𝑓(3) = 4. What is the approximation of 𝑓(2.8) found by using the tangent to the graph of 𝑓 at π‘₯ = 3?

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Video Transcript

For the function 𝑓, 𝑓 prime of π‘₯ is equal to a half π‘₯ plus three and 𝑓 of three is equal to four. What is the approximation of 𝑓 of 2.8 found by using the tangent to the graph of 𝑓 at π‘₯ equals three.

This question is asking us to use tangent line approximation, which is a method of approximating the value of a function at a given point by using the tangent to the graph of the function close to that point. This method assumes that there is a brief interval around the point at which the tangent has been drawn, for which the 𝑦-values on the tangent line give a good approximation to the 𝑦-values on the curve, that is, the values of the function itself. We can see from this simple sketch that there is indeed an interval around the point at which the tangent has been drawn, for which the vertical differences between the curve and the tangent are fairly small.

The formal result for tangent line approximation is as follows. If a function 𝑓 is differentiable at the point π‘₯ equals π‘Ž, then the equation of the tangent line, which can be used to find a linear approximation to the function, is 𝑙 of π‘₯ equals 𝑓 of π‘Ž plus 𝑓 prime of π‘Ž multiplied by π‘₯ minus π‘Ž. We know that our function 𝑓 is differentiable because we’ve been given its derivative 𝑓 prime of π‘₯ in the question. We’re told that we want to use the tangent to the graph of 𝑓 at π‘₯ equals three. So the value of π‘Ž in this question is three. The value at which we’re looking to approximate the function is 2.8. So the value of π‘₯ in the question is 2.8. 2.8 is fairly close to three. So it’s reasonable to use tangent line approximation.

Let’s substitute what we know into the question. If π‘₯ is equal to 2.8 and π‘Ž is equal to three, then the tangent line approximation of 𝑓 at 2.8, which we can write as 𝑙 of 2.8 is equal to 𝑓 of three plus 𝑓 prime of three multiplied by 2.8 minus three. Now, 𝑓 of three is given in the question; it is equal to four. 𝑓 prime of three can be found by substituting three into the expression given for 𝑓 prime of π‘₯. That’s a half of three plus three. And 2.8 minus three is equal to negative 0.2. So we have that 𝑙 of 2.8 is equal to four plus three over two plus three multiple by negative 0.2.

Now, we can think of three as six over two and then three over two plus six over two is equal to nine over two. We can also think of negative 0.2 as negative one-fifth, so we can convert to a fraction. Multiplying these two factions together will give us nine multiplied by negative one, that’s negative nine, over two multiplied by five, which is 10. So we have four plus negative nine over 10 or four minus nine-tenths. Now, as a decimal, nine-tenths is equal to 0.9. So we have four minus 0.9, which is 3.1. Using tangent line approximation then, we found that the value of the function 𝑓 of π‘₯ when π‘₯ equals 2.8 is approximately 3.1.

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