We want to compute the right hand and left hand derivatives as limits and check whether the function is differentiable at the point. So the first thing we're going to realize is we're using the definition as the limit as H approaches zero of F of X + H -, F of X all divided by H. So in this problem, when they're saying we want to find F of 2 + H, our X is really just going to be two. That's important because when we look at the left side and the right side, as H approaches zero, we need to know what equations we're using. So if we're going H approaches zero from the right side where X is 2, we're coming in this direction. So we're going to use our Y equal X -, 4 equation. So every time we see our X, we're going to stick in that 2 + H. So we're going to have 2 + H - 4 - F of two. We're going to stick 2IN when we see our X 2 - 4 all over H, so we get 2 + H - 4 - 2 + 4 all over H. The twos cancel, the fours cancel. H / H is one. So the limit as H approaches zero from the right side of one is just one. We're going to do the same thing approaching 2 from the left side because we want zero to get smaller and smaller and smaller from the left. So now we've got to use this X -, 2 ^2 - 2 equation. So the limit as H goes to zero from the left of 2 + H - 2 quantity squared -2 minus. We're going to stick to and for the X 2 -, 2 ^2 -, 2 all over H. So back here, this two and this -2 are going to cancel, leaving me just an H ^2 in the first quantity squared. And then we're going to have the -2. Then the 2 -, 2 is just going to cancel, leaving a Sierra squared -2. When we distribute out the negatives, we can see that the two -2 and +2 will cancel, leaving us an H ^2 / H, which is really just H So as H goes to zero from the left, we get 0 for that limit. So the limit from the left side is one is zero, and the limit from the right side is one. And the fact that zero and one are not equal means that it's not differentiable at that point.