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Numerade Educator

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Problem 59 Hard Difficulty

Find all points on the graph of the function $ f(x) = 2 \sin x + \sin^2 x $ at which the tangent line is horizontal.

Answer

$\left(\frac{\pi}{2}+2 \pi n, 3\right)\left(-\frac{\pi}{2}+2 \pi n,-1\right)$

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Video Transcript

Our goal here is to find all points on this function where the tangent line is horizontal and if a tangent line is horizontal than its slope will be zero. That means we're looking for all the points that have a derivative of zero. So let's find the derivative. But first I'm going to rewrite the function as to sign of X plus the sign of X quantity squared, which is what it means if we write science Codex. But it's a little bit more difficult to use the chain rule when it's written the science cortex. Okay, so now for the derivative, the derivative of To Sign Next is to co sign X and for the derivative of sign of X quantity squared will use the chain rule first, the derivative of the outside. We bring down the two and we raise sign next to the first and then the derivative of the inside. The director of the sign is co sign. Okay, We want to know where this equals zero so that we get the tangent lines to be horizontal. So let's set this equal to zero and notice that both terms have a to co sign next that we can factor out. Yes, we have to co sign X times one plus sine X is equal to zero, so we can set each factor equal to zero when we set to coast Line X equal to zero and then divide by two co sign X equals zero X would be pi over two or negative pi over two. And when we set one plus sign of X equal to zero, subtract one from both sides. Sign of X equals negative. One X would be negative pi over two, which is the answer we already got in the previous part. So we have two answers with pi over two and negative pi over, too. But don't forget that those air only the answers when you've gone around the unit circle one time you could go around again infinitely many times, and you could go in the other direction as well. So they're infinitely many answers that give you the same co sign and sign values so we'll write those as hi over to plus two pi n to Pai n represents going around the circle end times and could be one and could be too, and could be three and could be any integer and the other one would be negative. Pi over two plus two Pion. Same idea. Okay, The problem asks for the points on the graph, so that means it wants X coordinates and y coordinates. We have X coordinates. We need to find Why Coordinates. So let's do that next. So let's take our function. F of X equals two sine x plus the sign of X squared and let's find the Y Coordinate for Pi over two. So that would be two times. Sign of Pi over two, plus a sign of pi over two school. Where'd sign of Pi over two is one. So if two times one plus one squared, so we get three. And now let's find f of negative pi over two. Same idea. We have two times a sign of negative pi over two, plus the sign of negative pi over two squared and the sign of negative pi over to his negative one. So if two times negative one plus native one squared and that gives us negative one. So let's go back and write these ordered pairs. So we had high over two plus two pi n for the first set of X coordinates comma three for their Y coordinates. And then we had a negative pi over two plus two pi n representing the other answers and was positive one negative one negative one for the y coordinates Those air all the points where the horizontal where the tangent line is horizontal.