1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Question help please