• anonymous
I was just watching session 2 Clip 2, where the professor solves for the area of the triangle formed by any tangent line to y = 1/x and the y- and x-axes. While he finds the x-intercept of the tangent in a straightforward manner, using the equation y - y0 = -(1/(x0)^2) * (x - x0), from the fact that the derivative of y = 1/x is y = -1/x0^2, making the equation of a tangent line to y = 1/x at some point (x0,y0) what I listed above in point-slope form. But when finding the y-intercept, he takes a shortcut using the fact that y = 1/x is its own inverse function. Can anyone explain this shortcut
OCW Scholar - Single Variable Calculus
  • Stacey Warren - Expert
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  • chestercat
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  • phi
If you rename x to y and y to x, you end up with the same form of equation, and that equation will have the same value for its intercept (but for x).
  • anonymous
Let me use an example in physics to explain inverse function: We know a formula that converts F temperature to C temperature C = 5/9 (F-32) To find C, what we do is subtract 32 from F and then multiply by 5/9. To find F, we have to reverse the process. Multiply C by 9/5 and then add 32. Therefore, F = (9/5 * C) +32 Similarly, the inverse function of y = 1/x is x =1/y (replacing y with x and x with y) Now, you know the equation of the tangent line as y-1/xo = -1/xo^2 (x-xo) Interchange x and y, xo and yo Plugging the value y=0, you get y = 2yo In simple words, if y= 1/x you get x=2xo. If x = 1/y you get y = 2yo Hope you understand this.

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