Problem 8 Easy Difficulty

Find an equation of the tangent line to the curve at the given point.

$ y = \dfrac{2x + 1}{x + 2} $, $ (1, 1) $

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03:39

Daniel J.

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Video Transcript

suppose you want to find an equation of the tangent line to the curve. Why? Which is equal to two? X plus one over X plus two. At the .1, 1 to do this, we first find the slope of the tangent line. That is the derivative of the function Evaluated at the given .1, 1. Now by quotient rule we have white prime, that's equal to X plus two times the derivative of the numerator two, x plus one-. We have to express one times the derivative of the denominator which is expressed to this all over the square of the denominator. And then from here we have X plus two Times derivative of to Express one. That's just too minus two, X plus one times the derivative of X plus two which is one. And then this all over the square of X plus two. And simplifying this, we have two, X plus four minus two, X -1. This all over X plus two squared or this is just three over the square of experts to And so the slope of the tangent line at the .11 is given by that's dy over dx evaluated at 1 1. This is just three over one plus two squared. That's just 3/9 or 1/3. So this is the slope of the tangent line at 11. And then the next step would be to use the point slope formula of a line to find the equation of the tangent line. Now the point slope formula of a line states that the the equation of the line is just why minus? Why is that one? This is equal to the slope mm Of that line times X -X. Sub one. Since you already have our point Except one White 1 Which is just 11. And we have our slope, we found out To be won over three. Then the equation of the tangent line is just Why -1 That's equal to 1/3 times x -1. And simplifying this, we have why that's equal to 1/3, X minus 1/3 Plus one. Or that why is equal to 1/3 Plus 2/3. And so this is the equation of the tangent line at the point 11