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Find the points on a graph at which the tangent line is parallel

  1. Oct 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the points on the graph y=x^3/2 - x^1/2 at which the tangent line is parallel to y-x=3.


    2. Relevant equations



    3. The attempt at a solution
    First I found that the derivative of y=x^3/2 - x^1/2 is 1x.
    I then rewrote the other line as y = 3+x and found the derivative to be 1.
    Then I set 1x = 1, therefore x = 1
    Now I plugged it back into the original function and got y = 1^3/2-1^1/2 = 1
    So the point is (1,1).
    But I have no idea if I am anywhere near correct. Any help would be greatly appreciated.
     
  2. jcsd
  3. Oct 23, 2012 #2

    Mark44

    Staff: Mentor

    No it isn't. How did you get 1x for the derivative?
    Also, calculus questions should be posted in "Calculus & Beyond" not in the Precalc Math section. I am moving this thread.
     
  4. Oct 23, 2012 #3
    My apologies for the mispost. Also I do not know what I was thinking with that derivative. It should be 3x^1/2/2-1/2x^1/2 correct?
     
  5. Oct 23, 2012 #4
    Therefore, setting 3x^1/2/2-1/2x^1/2 = 1, I get x = 1/9, 1.
     
  6. Oct 23, 2012 #5

    Mark44

    Staff: Mentor

    Almost, and you need to use parentheses around the exponents if you write using ordinary text.

    The first term, (3/2)x^(1/2), in your derivative is correct, but the second one isn't.
     
  7. Oct 23, 2012 #6

    Mark44

    Staff: Mentor

    Here's the first part of what you wrote: 3x^1/2/2

    Using the precedence of operations, this is the same as if written ((3x^1)/2)/2

    Or, in LaTeX:
    $$ \frac{\frac{3x^1}{2}}{2} = \frac{3x^1}{4}$$
     
  8. Oct 23, 2012 #7
    The second part of the derivative, (1/2)x^(-1/2) = (1)/2x^(1/2) is incorrect?
     
  9. Oct 23, 2012 #8
    Ah, perhaps it is (1)/2√x
     
  10. Oct 23, 2012 #9

    Mark44

    Staff: Mentor

    (1/2)x^(-1/2) is correct, but it's not equal to (1)/2x^(1/2).

    This is also incorrect.

    According to the order of operations, (1)/2x^(1/2) means
    $$ \frac{1}{2}\cdot x^{1/2}$$

    Similarly, (1)/2√x means
    $$ \frac{1}{2}\cdot \sqrt{x}$$

    If you write these using inline text, this will do it: 1/(2x^(1/2)) or 1/(2√x). Now it's clear what's in the denominator.
     
  11. Oct 24, 2012 #10
    Regardless of my incorrect use of inline text. Did I solve for x properly when I came to x= 1/9 and 1?
     
  12. Oct 24, 2012 #11

    Mark44

    Staff: Mentor

    If you substitute either of these values (x = 1/9, x = 1) into the derivative, do you get 1 as the result?
     
  13. Oct 24, 2012 #12
    x = 1 does, x = 1/9 seems to give me -1.
     
  14. Oct 24, 2012 #13
    Clearly I am doing something very wrong.
     
  15. Oct 24, 2012 #14

    Mark44

    Staff: Mentor

    x = 1 is a solution. How did you get x = 1/9?
     
  16. Oct 24, 2012 #15
    (3x^(1/2))/2 - 1/(2x^(1/2)) = 1
    (3x^(1/2))/2 · 2x^(1/2) - 1/(2x^(1/2)) · 2x^(1/2) = 1 · 2x^(1/2)
    3x - x^0 = 2x^(1/2)
    3x - 1 - 2x^(1/2) = 0
    3u^2 - 2u - 1 = 0 where u = x^(1/2)
    (3u+1)(u-1) = 0

    Now 3u+1 = 0
    3u = -1
    (3u)/3 = -1/3
    u = -1/3
    x^(1/2) = -1/3
    x = (-1/3)^2
    x = 1/9
     
  17. Oct 24, 2012 #16

    Mark44

    Staff: Mentor

    So x1/2 = -1/3, which means there is no solution for x. There is no real number whose (principal) square root is negative.
     
    Last edited: Oct 24, 2012
  18. Oct 24, 2012 #17
    Ah. So the only solution is x = 1, which I plug back into the original equation y = x^(3/2) - x^(1/2), and gives me 0? Therefore, the point at which the tangent line is parallel to y-x=3 is (1,0)?
     
  19. Oct 24, 2012 #18

    Mark44

    Staff: Mentor

    Looks good.
     
  20. Oct 24, 2012 #19
    Thank you so much.
     
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