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Homework Help: Find the gradient of the curve

  1. Dec 18, 2004 #1
    Find the gradient of the curve [tex]y=\frac{5x-4}{x^2}[/tex] at the point where the curve crosses the x-axis.
    After I differentiating the equation, I got [tex]-\frac{5}{x^2} + \frac{8}{x^3}[/tex] (it might be wrong). Now what do I do?
     
  2. jcsd
  3. Dec 18, 2004 #2
    Well, dy/dx gives you the gradient at x. The curve crosses the x-axis when y = 0. You've worked out the derivative correctly though.
     
  4. Dec 18, 2004 #3
    I thought so too. But I couldn't get the right answer.
     
  5. Dec 18, 2004 #4
    What answer did you get?
     
  6. Dec 18, 2004 #5
    [tex]3\frac{1}{8}[/tex]
     
  7. Dec 18, 2004 #6
    Can you post your working? I got dy/dx = 7.8125. Also, what value did you get for the x intercept?
     
    Last edited: Dec 18, 2004
  8. Dec 18, 2004 #7
    To get x I must substitue y=0 into [tex]y=\frac{5x-4}{x^2}[/tex] right?
    Btw the answer my book gives me is the same as yours
     
  9. Dec 18, 2004 #8
    Yes, then solve for x.

    y = (5x - 4)/x^2

    => (5x - 4)/x^2 = 0

    Need to solve for the numerator being equal to 0, so x = 4/5.
     
  10. Dec 18, 2004 #9
    Thats great. Then I plugged in 4/5 into
    [tex]-\frac{5}{x^2} + \frac{8}{x^3} \rightarrow -\frac{5}{\frac{4}{5}^2} + \frac{8}{\frac{4}{5}^3}[/tex]
    Then I solve and get [tex]3\frac{1}{8}[/tex]
     
  11. Dec 18, 2004 #10
    Oh now I know why I didn't get the answer. I calculated a part wrong. Sorry. Thank you for your time.
     
  12. Dec 18, 2004 #11
    No problem :smile:.
     
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