Solve Q: Evaluate h(1) & h'(1)

[camboguy]

ok iv tried this like 2 times and I am 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) I am 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 I am not sure what my cord nets are to do the change in y over change in x, and the point that I am so posed to use to find the derivative of h(1)

 

[HallsofIvy]

ok iv tried this like 2 times and I am 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) I am 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 I am not sure what my cord nets are to do the change in y over change in x, and the point that I am so posed to use to find the derivative of h(1)

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.