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Homework Help: Question help please

  1. Sep 6, 2007 #1
    ok iv tried this like 2 times and im still not geting it

    Q> suppose that h(x)=f(x+1) and that the line tangent to the graph of f at the point (2,f(2)) is described by the equation y = 3x-4.

    a) evaluate h(1)
    b) evaluate h'(1)

    heres what i know and did.

    i under stand that for h(1) i just plug in 1 for h(x) = f(x+1) and get f(2)

    but for h'(1) im so posed to find the slope at h(1) and so i figured that since y=3x-4 is the tangent line that touches h(x)=f(x+1) at the points (2,f(2)), the functions are equal and have the same (x,y) and so i was trying to find the slope but im not sure what my cord nets are to do the change in y over change in x, and the point that im so posed to use to find the derivative of h(1)
  2. jcsd
  3. Sep 6, 2007 #2


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    Science Advisor

    Okay, h(x)= f(x+1) so h(1)= f(1+1)= f(2). You also know, by the chain rule, that h'(x)= f '(x+1) since the derivative of x+ 1 is 1.

    You are told that the tangent line to y= f(x) at (2, f(2)) is given by y= 3x-4. Of course, the tangent line passes through the point. y= 3(2)- 4= 2 so (2, f(2)) must be (2,2). h(1)= f(2)= 2. Also the slope of the tangent line, 3, is the derivative at that point so f '(2)= 3 and then h'(1)= f '(2)= 3.
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