If $ f $ is the function whose graph is shown, let $ h(x) = f(f(x)) $ and $ g(x) = f(x^2). $ Use the graph of $ f $ to estimate the value of each derivative,
(a) $ h'(2) $
(b) $ g'(2) $
April 4, 2021
If x2 1 xy 1 y3 ? 1, find the value of y999 at the point where x ? 1.
All right. So for this function, we have a graph and you should definitely use the graph provided by the book because my graph is just a rough sketch and I might not be as accurate. So we want to find HPE prime of two and we're told that h of X is f of f of X and that's a composite function. So we're going to use the chain rule to find its derivative. We take the derivative of the outside, which would be f prime of f of X and multiply that by the derivative of the inside, which would be a prime of X and then we substituted to when they're to find h prime of to. So everywhere we haven't X, we replace it with the to. Now we take a look at our graph to get some information about what f of two is. So you look for where X equals two. And then where is that on the graph? That's at a height of one so we can substitute a one in there. And now we have f prime of one times f prime of to so f prime of one. The derivative of one is the slope to the tangent line. The slope of the tangent line at that point in F prime of two, is the slope of the tangent line at that point. So let's find X equals one. And let's imagine a tangent line drawn in there. Okay, I could imagine that that line has a slope of roughly negative 1/2 so F prime of one is about negative 1/2. It might not be exactly that. That's my rough estimate. And then same idea for F prime of to You have X equals two. We have the point on the graph and then imagine the tangent line drawn there. Maybe that line has a slope of negative, too. So have negative 1/2 times negative, too. So we are. Answer would be one, and that's just a estimate. It might not be exactly one. Okay, we have a similar thing to do. For part B. G of X is f of X squared. Another composite function. Use the chain rule to get G prime of X, and that would be f prime of X squared times, the derivative of X squared, So times two X now for G prime of two. We substitute to infer X. We've f prime of two squared, and that would be a prime before times two times two. So now let's look at the graph. Find X equals four and find where that is on the function. And then let's approximate the slope of the tangent line there. So suppose that's about positive two. So we have two times, two times two, so we get approximately eight.