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Derivatives and Slopes

  1. Jan 30, 2014 #1
    Hello,

    For this question I managed to find an answer but I am not sure if what the question means is to plug in the x-value and find the slope, or first find the derivative of the function, and THEN solve for the slope using f'(x)=10x+5h-2 (what I got for the derivative of the function). If someone could help explain the question I would appreciate it.

    1. The problem statement, all variables and given/known data
    For each function, use the definition of the derivative at a point to find the slope of the tangent at the indicated point.

    2. Relevant equations

    For point (2, 17): [tex]f(x) = 5x^2-2x+1[/tex]

    3. The attempt at a solution
    [tex]f'(x) = lim_h→0 \frac{f(x+h)-f(x)}{h}[/tex]
    [tex]= lim_h→0 \frac{[5(2+h)^2-2(2+h)+1]-[5(2)^2-2(2)+1]}{h}[/tex]
    [tex]= lim_h→0 \frac{[5(4+4h+h^2)-4-2h+1]-[20-3]}{h}[/tex]
    [tex]= lim_h→0 \frac{20+20h+5h^2-3-2h-20-3}{h}[/tex]
    [tex]= lim_h→0 (18+5h)[/tex]
    [tex]= 18 + 5(0)[/tex]
    [tex]= 18[/tex]
     
  2. jcsd
  3. Jan 30, 2014 #2

    Mark44

    Staff: Mentor

    So f(2) = 5(2)2 - 2(2) + 1 = 20 - 4 + 1 = 17
    In the line above you aren't finding f'(x), which is implied by the line above it. What you are doing is finding f'(2). Your first equation should have been
    $$f'(2) = lim_h→0 \frac{f(2+h)-f(2)}{h} $$
    Aside from what I already said, this looks fine. What your answer means is that at the point (2, 17), the slope of the tangent line is 18.
     
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