 In this video we provide the solution to question number eight for a practice exam number four for math 1210 in which case we're asked to use the linearization of the function f of x equals square of x at the point a equals 64 to approximate the square root of 64. So remember how the linearization is computed. So L of x is equal to f prime at a times x minus a plus f of a. So let's compute each of these things one by one. Let's start off with the derivative. So the derivative here f prime of x by the usual power rule is going to be one over two times the square root of x. We need to evaluate this at the a value 64. So f prime at 64, we end up with one over two times the square root of 64. The square root of 64 is going to end up being just of course eight. You times that by two, you're going to end up with one sixteenth. So that's the slope of this line. Then we're going to get x minus while we have to compute just the a value, which is 64. And then you're going to add to it f of 64, which against the square root of 64, which is going to be eight. This is what we have to do. So we need to compute L of 65. That's what we're tasked with doing here. So we're going to get one sixteenth times 65 minus 64 plus eight for which 65 take away 64 gives us one. So we're going to end up with one sixteenth plus eight like so. So we have to add these fractions together. Of course, if you take eight times 16, that's 128. So we're going to get one sixteenth plus 128 over 16. So adding those together, we see that the correct answer is going to be B 129 over 16. That's going to be the estimate of the square root of 65 given by this linearization method.