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Numerade Educator

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Problem 27 Medium Difficulty

If $ f(x) = 3x^2 - x^3 $, find $ f'(1) $ and use it to find an equation of the tangent line to the curve
$ y = 3x^2 - x^3 $ at the point $ (1, 2) $.

Answer

$f^{\prime}(1)=3 \quad y=3 x-1$

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November 4, 2020

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Video Transcript

So we've got the function F of x equals three, X squared minus execute. Um Our ultimate goal is to find the equation of the tangent line to this function at the point on to. Um So it's kind of a hint here, we first need to find f prime of one which is the derivative at the point X equals one, which is where this point is that we want to find the tangent line at. So first to find F prime of one, we need to find the derivative function. So F prime of X. So let's do that using the power rule to gets multiplied out front. So two times three is six X. We subtract one from the exponents. So we just have an exponents of one derivative of X cubed three comes outfront and then we have X squared, decrease the power by one. So this is our derivative function. So let's find what is the derivative or what is the slope At? X. equals one. So let's plug in one for X. So six times one -3 times one squared Which is 6 -3, which is equal to three. So we know that the slope when x is equal to one, the slope is three. So this is our M. And uh equation of a tangent line, we know a tangent line is a straight line, right? And the equation of a line always looks like why it goes mx plus B. And we already found out what M is right. We just discovered that's three and we know a point on the line, Right? The standard line is at the .1, 2. So it cross through that point. Um So we know an X coordinate X. Is one and we know why is too, so we can find B by plugging this in, right? So let's plug in those numbers. So two is equal to three times 1 Plus B. So subtracting three from both sides, we get -1 is equal to be. So the final answer, the equation of our tangent line is why equals MX right? So three x Plus B -1. So this is the equation of our tangent line at that.