Power rule

  • #1
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Homework Statement



Find two straight lines that are perpendicular to y=0.25x and tangent to the curve f(x) = 1/x.

Homework Equations



y=0.25x
f(x) = 1/x.

The Attempt at a Solution



What I did was equate y and f(x) and determined when they equal which is 2 and -2. The points are (2, 1/2) and (-2,-1/2).

Now, I took the derivative of f(x) and got -1/x^2. Would I use this to find the slope at the two points? Then I could make the equation.
 

Answers and Replies

  • #2
Dick
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If the lines are perpendicular to y=(1/4)*x then what should the slopes of the tangent lines be? The slopes of the tangent lines should also be f'(x), right?
 
  • #3
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If the lines are perpendicular to y=0.25x, what is their slope?
 
  • #4
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If the lines are perpendicular to y=(1/4)*x then what should the slopes of the tangent lines be? The slopes of the tangent lines should also be f'(x), right?

The slope would be -4, no?

I've made the two equations of y = -4x + 17/2 and y = -4x -17/2.
 
  • #5
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The -4x is right, but your y-intercept is wrong. Now that you have the slope of the lines, what does the relationship have to be between that slope and f'(x) in order for the lines to be tangent to f(x)?
 
  • #6
Dick
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The slope would be -4, no?

I've made the two equations of y = -4x + 17/2 and y = -4x -17/2.

Yes, the slope should be -4. But how did you get those two line equations?
 
  • #7
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Yes, the slope should be -4. But how did you get those two line equations?

I equated y and f(x) to see when they intercept. I got 2 and -2. I put these two into f(x) to get the y cordinate so I can make the equation as I have the slope.
 
  • #8
Dick
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I equated y and f(x) to see when they intercept. I got 2 and -2. I put these two into f(x) to get the y cordinate so I can make the equation as I have the slope.

I'm still not totally clear what you are doing. But if you know the slope is -4, then f'(x) should be -4, right? What are the possibilities for x?
 
  • #9
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I'm still not totally clear what you are doing. But if you know the slope is -4, then f'(x) should be -4, right? What are the possibilities for x?

Ahhh, I equate the slope to -4. The possibilities of x are 1/2 and -1/2?
 
  • #10
Dick
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Yes! If you know x=1/2 or -1/2, then you know y. So now you know x and y and the slope. Pretty easy, right?
 
  • #11
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Yes! If you know x=1/2 or -1/2, then you know y. So now you know x and y and the slope. Pretty easy, right?

y = just 2 and -2.

Yes, sir. Dang my silly mistake :[. Thank you!
 

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