Suppose f is differentiable on (-∞,∞) and the equation of the ...

Question

Suppose f is differentiable on (-∞,∞) and the equation of the line tangent to the graph of f at x = 2 is y = 5x -3. Use the linear approximation to f at x = 2 to approximate f(2.01).

Accepted Answer

Hi everyone, let's take a look at this practice problem dealing with linear approximation. This problem says assume that J is a function that is differentiable across all real numbers, with a tangent line at X equal to 0.5 given by Y is equal to 3 x plus 1. Using the linear approximation to j at x equal to 0.5, calculate an approximation for J of 0.51. We're given four possible choices as our answers. For choice A, we have 1.53. For choice B, we have 1.47. For choice C, we have 2.53, and for choice D, we have 2.47. Now we're asked to calculate an approximation for J of 0.51 using a linear approximation. So in order to use the linear approximation, we will need to find the value of our function J, as well as the value of the derivative of our function J at X equal to 0.5. So we'll start off first by trying to find J of 0.5. So here we're trying to find the derivative of our function at X equal to 0.5. And we're told that the equation for the tangent line at X equal to 0.5 to our function J is given by Y is equal to 3 X + 1. Now recall that the derivative of our function at that point is equal to the slope of our tangent line. So in this case, J of 0.5 is equal to the slope of our tangent line, which is 3. We also need to determine the value of J of 0.5. And here we call that our tangent line intersects our function J at the point of X equal to 0.5. So that means since they intersect, they must have the same value. That means that J of 0.5 is equal to 3 multiplied by 0.5. Plus one. Which, when we evaluate this expression, is equal to 2.5. So now we have all the information that we need to use our linear approximation. So recall for the linear approximation, we have L of X. is equal to J A. Plus J A multiplied by the quantity of X minus A. So, in our case here, A is going to be equal to 0.5. And X is going to be equal to 0.51. So we're looking for J of 0.51. So, substituting these values into our expression, we will have L of 0.51. is equal to J of 0.5, which we found to be 2.5. Then we'll have plus of 0.5, which we found to be 3. That gets multiplied by X minus A, which is going to be 0.51 minus 0.5. So, simplifying this expression, this is going to be equal to 2.5. Plus 3, multiplied by 0.01. Which is going to be equal to 2.5. Plus 0.03, which is equal to 2.53. This here is the answer that we were looking for for this problem, and we can compare that to our answer choices, we see that the correct answer choice corresponds to answer choice C. So I hope that this explanation has been useful, and I'll see you in the next video.

Suppose f is differentiable on (-∞,∞) and the equation of the line tangent to the graph of f at x = 2 is y = 5x -3. Use the linear approximation to f at x = 2 to approximate f(2.01).

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