Critical points Find the critical points of the function ƒ(x) ...

Question

Critical points Find the critical points of the function ƒ(x) = sinh² x cosh x.

Accepted Answer

Welcome back, everyone. In this problem, we want to determine the X coordinates of the critical points of the function H X equals shine cubed X multiplied by cache X. Now we can determine the x coordinates of the critical points by first differentiating our function H of X, setting it to zero, and solving for X. So let's find the derivative of H of X. Now notice H of X is a product. It's shine cubed X multiplied by cache X. So to differentiate a product we can use the product rule. Now by the product tool. OK. Recall that if we have a function y equals UV. Then the derivative of Y will be equal to the derivative of u multiplied by V plus u multiplied by the derivative of V. In other words, we differentiate both terms and then multiply them by the opposites. So in this case, if we let U be equal to shine cubed X. And V be equal to cache x. Then the derivative of u with respect to X will be equal to 3 times squared x multiplied by cash x. And the derivative of V is going to be sin x. So now we can multiply both to find the derivative. Because this tells us then that the derivative of H of X. Assuming or letting H of X be Y would be equal to 3 squared X X. Multiplied by cash X OK. Plus shine cubed X multiplied by sin x. Now, if we were to simplify, then we'll get 3 shine squared X, squared X, OK. Plus shine. X to the 4th, OK? So shine x to the 4th. And now, you probably noticed that both terms have shine squared X in common. So if we factor out shine squared X, OK, then we're going to have 3 cash squared X. Plus shine squared X in or bracket. So now we've found the derivative of H of X. Now at the critical points, like we said before, OK. Then Our derivative is going to be equal to 0. So this would imply then that we would set shine square x multiplied by. Let me just direct it the way it was before here, multiplied by 3 squared X plus sine quad X. To be equal to 0. And if this expression equals 0, then this means that shine square x. Will be equal to 0, OK. Or OK, or, uh, 30. Squared X plus square x will be equal to 0. So let's solve for X in both scenarios. Now if square X equals 0, then X will be equal to 0. On the other hand, to solving our second problem, recall that shine 2 X equals 2x minus 1. So we could rewrite it. Let me just put a line here. We could rewrite it as 332 X. OK. Plus cash square x. -1 equals 0. Now if we expand, we get 4 square x minus 1 to be equal to 0, and if we add 1 to both sides and divide by 4, then we get cost square x to be equal to 1/4, OK, and we could go ahead and sell for cash, but hold on one minute, recall again that cash square x. Is going to be greater than or equal to 1 for all real numbers of X. OK. Therefore, there is no real solution for this equation. In other words, cost squared X cannot be equal to 14th. And if that's the case, that means the only real critical point is that x equals 0. Therefore, we can say that the critical point. OK, a critical point. Of HF X. Is at X equals 0. Thanks a lot for watching everyone. I hope this video helped.

Critical points Find the critical points of the function ƒ(x) = sinh² x cosh x.

Key Concepts

Hyperbolic Functions

Critical Points

Product and Chain Rule in Differentiation

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