How Is the Value of f(5) Determined Given Parallel Tangent Lines?

[syeh]

Homework Statement

The line x-2y+9=0 is tangent to the graph of y=f(x) at (3,6) and is also parallel to the line through (1,f(1)) and (5,f(5)). If f is differentiable on the closed interval [1,5] and f(1)=2, find f(5)

A) 2
B) 3
C) 4
D) 5
E) None of these

The correct answer is (C) 4

The Attempt at a Solution

So I know the tangent line to (3,6) and f(5) have the same slope:

x-2y+9=0
2y=x+9
y=.5x+4.5

So, the slope is .5 at f(1), f(3), and f(5)

Now, I need to find the point at f(5). I do not know how to do this

 

[janhaa]

1) The slope of the first line is 1/2

2) The slope of the line is passing through (1,2) and (5, f(5)). Then use the definition of slope:

[tex]1/2=\frac{f(5)-2}{5-1}[/tex]

and solve for f(5).