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Determine the equation of the tangent line at the given $x$ -coordinate.$$f(x)=x e^{-2 x}, x=0$$

$y=x$

Algebra

Chapter 4

Exponential and Logarithmic Functions

Section 4

The Derivative of the Exponential Function

McMaster University

Harvey Mudd College

Baylor University

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we have another tangent line problem, and any time you see the phrase tangent line, just think, Oh, I better get a point and a slope and then the answer is pretty straightforward. And in this problem, they tell you the X coordinate of the point and then to find the slope. It's the derivative plugging in the X coordinate that they give you in both those cases is zero. So as you look at the equation X e to the negative two x eso That's the equation we're looking at. Well, if you plug in zero and for both of these excess, well, zero times anything will give me zero. So it is zero times one, but the like coordinate is zero. So what you're left with doing is fighting the derivative, which is your product rule. So you take the derivative of the left side. You leave the right side alone. Plus, now you leave the left side alone times the derivative of the right side. And that's your chain rule. The only thing is is now we're tasked with plugging in zero again. Well, what we're left with is e to the zero power because name to time 00 and then plus zero times eat of the zero again times negative, too. But that doesn't matter, because zero times anything is nothing and eat of the zero power equals one. So what's the equation? Um, I like to do point slope form, but actually this finds you the y intercept when x zero eso. It's actually really fast if you just write as y equals one X because the line except has to be zero in the problem on usually. What math teachers will do is that there's no needed to write times one in a problem. So just why equals X on that's in point. So form while he goes, I want X Plus zero is acceptable to.

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