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Need help with solving the slope of a tangent line

  1. Jun 4, 2012 #1
    1. The question is What is the slope of the line tangent to f(x)=2x^2-x-7 at x=-1



    2. Must be solved using derivatives



    3. So basically i know the equation (f(x+deltax)-f(x)/deltax)
    So i plug in the problem and i get (2(x+deltax)^2-(x+deltax)-7)-(2x^2-x-7)
    Now i know what the answer is to the next step (2(x^2+2xdeltax+deltax^2)-(x+deltax)-7)-(2x^2-x-7) but i do not understand this step. I only have the answer because this problem is from a video, i just have no clue how they got there. I also do not know how to do the steps after. If someone can explain the algebra they used to do these steps (be very detailed) i will be very happy.
     
  2. jcsd
  3. Jun 4, 2012 #2
    Do you know differentiation and or the concept of limits? This basically is based on the definition of derivative which gives the slope of the curve as,

    [tex]Slope = f'(x) = \lim_{h\to 0} \frac{f(h+x) - f(x)}{h}[/tex]

    This f'(x) is what you are trying to find. Your method of delta x would work if it were a straight line, but since it is the curve, you bring f(x+Δx) closer and closer to f(x) and as Δx tends to 0, you get the derivative.
     
  4. Jun 4, 2012 #3

    Mentallic

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    So [itex]f(x)=2x^2-x-7[/itex] and thus [itex]f(x+\Delta x)=2(x+\Delta x)^2-(x+\Delta x)-7[/itex].

    So plugging that into the formula [tex]\frac{f(x+\Delta x)-f(x)}{\Delta x}[/tex] gives us [tex]\frac{2(x+\Delta x)^2-(x+\Delta x)-7-(2x^2-x-7)}{\Delta x}[/tex]

    Now let's simplify the numerator:

    [tex]2(x+\Delta x)^2-(x+\Delta x)-7-(2x^2-x-7)[/tex]
    [tex]=2(x^2+2x\Delta x +(\Delta x)^2)-x-\Delta x - 7 -2x^2 +x+7[/tex]

    [tex]=2x^2+4x\Delta x +2(\Delta x)^2-x-\Delta x -7-2x^2+x+7[/tex]
    Cancelling all like terms, we end up with:
    [tex]4x\Delta x + 2(\Delta x)^2-\Delta x[/tex]

    Now, can you finish off the problem?
     
  5. Jun 4, 2012 #4
    This is where I am having problems. I do not understand how you went from [tex]2(x+\Delta x)^2-(x+\Delta x)-7-(2x^2-x-7)[/tex]
    to [tex]=2(x^2+2x\Delta x +(\Delta x)^2)-x-\Delta x - 7 -2x^2 +x+7[/tex]
    And then to the next step. Can you explain this step in a little more detail?
    I feel like it is just simple algebra, im just forgetting what I have already learned
    Everything else (the cancling and plugging in) in understand.
    Also, i am teaching myself calc after two years of not taking a math course so i am rusty :P
     
  6. Jun 4, 2012 #5

    Mentallic

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    It's just some basic rules of algebra.

    [tex](a+b)^2=a^2+2ab+b^2[/tex] because [tex](a+b)^2[/tex][tex]=(a+b)(a+b)[/tex][tex]=a(a+b)+b(a+b)[/tex][tex]=a^2+ab+ba+b^2[/tex][tex]=a^2+2ab+b^2[/tex]

    and the second rule is something similar to

    [tex]-(a-b+c)=-a+b-c[/tex]
     
  7. Jun 4, 2012 #6
    I feel dumb now. I knew it was something easy haha.
    Answers -5 thank you guys
     
  8. Jun 4, 2012 #7
    Take the derivative of the function : F(x)= f(x)=2x^2-x-7
    of F'(x)=4x-1
    Thus,
    -1=4x-1
    0=4x
    as such, the slop occurs when the x-coordinate is zero.
     
  9. Jun 4, 2012 #8

    Mentallic

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    Wrong.

    Right.
     
  10. Jun 5, 2012 #9

    Mark44

    Staff: Mentor

    Aside from the answer below being wrong, the OP's problem was to find the derivative using the limit definition, so your answer above would probably have gotten no credit.
     
  11. Jun 5, 2012 #10
    @ Mark

    With all due respect, if I set the derivative equal to slope and solve for x, I would get the coordinate at which it occurs? Ohh your right, checked on the the calculator that was the wrong answer. What is the error in my answer?
     
  12. Jun 5, 2012 #11

    Curious3141

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    The question asks for the slope of the tangent line at x = -1. That's simply f'(-1) = 4(-1) - 1 = -5.

    Not for the x-coordinate for which the slope of the tangent line is -1. Here you have to solve f'(x) = -1, as you did.
     
  13. Jun 5, 2012 #12
    Alright thanks.
     
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