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2 Find the points on the graph of f(x) = 8x3 4x2 2x + 1at which the tangent line is horizontal...

Question

2 Find the points on the graph of f(x) = 8x3 4x2 2x + 1at which the tangent line is horizontal

2 Find the points on the graph of f(x) = 8x3 4x2 2x + 1at which the tangent line is horizontal



Answers

Find the points on the graph of $y=2 x^{6}-x^{4}-2$ at which the tangent line is horizontal.

Okay, so we're going to find all the points where we have a horizontal tangent line for this function and a horizontal tangent line just means that are derivative function is equal to zero. So we need to figure out where derivative function is equal to zero. So let's go ahead and write out our derivative function. It's going to be equal to the derivative of the first term. And I'm just using the product rule here. So the derivative of the first term is the derivative of X plus two which is just equal to one, then multiplied by the second term. So x squared minus two, X minus eight. And then plus the first term multiplied by the derivative of this second um term here or function if you want to call it a function. Um And so that's going to be the derivative of X squared which is two x minus the derivative of two X, which is two and then minus the derivative eight, which is equal to zero. So it's just gonna be two x minus two. So this derivative function is equal to x squared -2 x -8 and then plus two X squared Plus or -2 X. And then plus four x and minus four. And so this simplifies to what we have three X squared. We're going to have gonna have to negative two. X is in one positive for X. So those are actually all going to cancel we're gonna have no X terms and then we're gonna have negative eight minus four which is negative 12. And so now that we have our functions derivative, we just want to set this equal to zero and see what values we get for X. So we have zero is equal to three, X squared minus 12. 12 is equal to three X squared, divide both sides by three. We had X squared is equal to four, so then X here would be equal to plus or minus two. And so the two places where we're gonna have horizontal tangent lines are going to be at X is equal to two, and X is equal to negative two.

For the given problem, let's find the slope of the given function. So we have f of x equals eight x cubed -4 people. So we see that when x equals 1/2 the slope that we end up getting right here, That's going to equal 1/2. So the slope is going to be we have a negative three is our value, and the slope is going to be um six would be our slope. Um so if we have y equals six, X must be, you know that we end up getting a negative three when we plug in one half, so it's going to be six equals six times 1/2 is three. Down here is a -3. So we end up getting a negative six and that ended up being sure enough, the tangent line at the given point of the function.

The underlying theme in this problem is that we're finding the values four X. I guess I should mention that way. So I don't know what C is. But if we have a horizontal tension, that means that the slope is equal to zero. So as you're looking at this problem, what I'm gonna do is take the derivative of extra the fourth minus eight X squared plus three and set the derivative equal to zero. That z what we gotta do. So what's the derivative Will bring the four in front. It's now to the third power because you subtract one from your exponents two times eight is 16 next to the first and set that equal to zero. So now you can do this factor out a four x you're left with X squared minus four eso When you use a zero probably property, you can say Okay, well, ex convict zero. And I don't care what anything else is because if x zero we're going to get zero or x squared can equal four because that makes this piece zero on anything. Times zero would be zero. That means X could be plus or minus two. But If you pay close attention to the directions, it's asking for the points on the curve. So what that means is we have to take these values we found and plug them back into the original problem. Now plugging in zero is pretty easy because you get zero minus zero plus three. So that means the ordered pair of 03 The Y coordinate is one answer. Now these other ones. They're a little bit more complicated because if you plug in to 248 16 to to the fourth Power 16, just do some scratch work here. Two squared would be four times eight would be 32 so or at negative 16. So far, plus three gives me negative 13 to negative. 13 is another point on the curve. Um, so let's do the same thing with negative, too. Well, negative to to the fourth Power. Still positive. 16. Negative two to the second power. Still positive, so it's still negative. 32 s o. You know, you can use a calculator to verify, but it's gonna be the same y value on that one as well. Eso these air your three points 032 negative 13 and native to native 13

For this problem we are asked to determine the points at which the graph of the function F of X equals four X minus two. Over X squared has a horizontal tangent line. A horizontal tangent line means that f prime of X equals zero. So first of all we need to find f prime of X. So we can find that Using the question rule we'll have four times X squared minus two X times four X minus two, Divided by X to the power of four. Now we can expand out the numerator there. Or actually I'll say that, first of all we note that a fraction is only ever equal to zero if the numerator equal zero. So we need that numerator to equal zero. And I'll also expand out the numerator. So we'll have four X squared minus eight X squared plus four X has to equal zero. Which in turn means that we need negative four X squared plus four X two equals zero. Can factor out a negative four Or negative for acts, I should say to get X -1. So we can see that we'll have a horizontal tangent line when X equals zero and when X equals one


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