# Power rule

1. Jan 14, 2010

### fghtffyrdmns

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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.

2. Jan 14, 2010

### Dick

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. Jan 14, 2010

### GunnaSix

If the lines are perpendicular to y=0.25x, what is their slope?

4. Jan 14, 2010

### fghtffyrdmns

The slope would be -4, no?

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

5. Jan 14, 2010

### GunnaSix

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. Jan 14, 2010

### Dick

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

7. Jan 14, 2010

### fghtffyrdmns

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. Jan 14, 2010

### Dick

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. Jan 14, 2010

### fghtffyrdmns

Ahhh, I equate the slope to -4. The possibilities of x are 1/2 and -1/2?

10. Jan 14, 2010

### Dick

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. Jan 14, 2010

### fghtffyrdmns

y = just 2 and -2.

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