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# At what point on the curve $y = \sqrt {1 + 2x}$ is the tangent line perpendicular to the line $6x + 2y = 1?$

## Since the question asks for a point and not just a value of $x,$ let's get a valueof $y$ to finish off. $y=\sqrt{1+2(4)}=3,$ so the point we're looking for is $(4,3)$ .Finito. QED

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Chanse S.

November 15, 2020

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### Video Transcript

in this problem. We're looking for the point where the Tanja line is perpendicular to the given line and perpendicular lines have opposite reciprocal slopes. So let's find the slope of the given line. Let's take this equation and isolate why. So we'll subtract six x from both sides and divide by two. So the slope of this line is negative. Three. So the perpendicular slope would be positive 1/3. So we're looking for the point where the Tanja line has a slope of 1/3. That means we want the derivative to equal 1/3. So let's find the derivative. I'm going to rewrite the function as one plus two x to the 1/2 power and then I'm going to use the chain rule. Why Prime equals 1/2 times one plus two x to the negative 1/2 power that takes care of the first part of the chain rule times the derivative of the inside, which is to now we can multiply the 1/2 of the two that was canceled. And now we want this derivative to be equal to 1/3 so this derivative could be written as one over the square root of one plus two x, and it equals 1/3. Let's take the reciprocal of both sides, and we have a square root of one plus two. X is equal to three. Now it's square both sides, and we have one plus two. X equals nine. Subtract 12 X equals eight and divide by two. X equals four. Okay, this is the X coordinate of the point. We also want the Y coordinate of the point, and the Y coordinate would be what we get when we take four and substituted into the original equation. So we have one plus two times for inside a square root square root of nine is three. So the point we're looking for is the 0.0.43

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