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Find an equation of the tangent line to y = f(x) at x = 1. f(x) 3e^x

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@zepdrix hi can u help for a minute i know how to do this since you showed me yesterday how..but i'm stuck on something, so can u help?
@Mertsj can u help?
sup? c: what part confusing you?

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okay so i tried doing this one here is what i got so far i found the derivative of 3e^x and i got d/dx (3e^x) = 3(d/dx (e^x)) = 3e^x so now I filled in the 1 into 3e^x and I got 3e so i want to put it into y=mx + b and i got stuck there
Yah it's a weird looking number, so I can see how it's a little confusing. So you calculated the slope, \(\large m\). \(\large m=3e \qquad \rightarrow \qquad y=(3e)x+b\)
To solve for \(\large b\) remember that we needed a coordinate pair that we could plug into this. Where can we get a coordinate pair from? :o
hmmm 1 in the original question that would be our x and for y we would put 1 into 3e^x
k sounds good.
okay so our coordiantes will be (1, 3e) ? since when you plug in 1 into 3e^1 you get 3e
so 1 = (3e)x + b ?
wait i meant 1 = 3e(1) + b
\[\huge \left(\color{orangered}{x},\;\color{royalblue}{y}\right) \qquad = \qquad \left(\color{orangered}{1},\;\color{royalblue}{3e}\right)\]Plug those into here,\[\huge \color{royalblue}{y}=(3e)\color{orangered}{x}+b\] You didn't quite get them plugged in correctly. Maybe the colors will help you match them up.
ohhh okay hold on
so 3e = 3e(1) + b?
okay then i solve for be and i'm done right?
Yes, solve for b. Then rewrite your final equatoin with y, x, m and b. Good job c:
so i fot 3e = 3e(1) + 0
yes, but you want to write it as an equation involving x and y. We no longer want the coordinate pair plugged in for our final answer.
ok how would i do that?
Remember we were trying to come up with an equation for the tangent line. It will be of the form \(\large y=mx+b\). We ONLY want to fill in the missing \(\large m\) and \(\large b\). We had plugged in our coordinate pair to help us find \(\large b\), but now we don't need that coordinate pair.
x=x y=y :d

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