Question

Find all values for c in the intercal (1,4) so that the slope of the tangent line at (c,f(c)) equals the slope of the secant line through (1,1/3) and (4,2/3) where f(x) = x/(x+2) 1 </ x </ 4 (<-- that is suppose to be the less than or equal to sign)

I'm not sure how to do this. I think this involves the Mean Value Theorem but I'm not sure where I should begin.

Answer 1

This is indeed the MVT.
The secant line through (1,1/3) and (4,2/3) has slope 1/9. The MVT says that in the interval (1,4), there is some c where f'(c) = 1/9.

f'(x) = 2/(x+2)^2
So, we want c such that
2/(c+2)^2 = 1/9
(c+2)^2 = 18
c+2 = ±3√2
c = -2±3√2
We want c to be in (1,4), so that means we want c = -2+3√2 = 2.24

So, as the MVT says, there is such a c.
f(-2+3√2) = (-2+3√2)/(3√2) = 1 - √2/3
Thus, the tangent line is

y-(1-√2/3) = 1/9 (x-(-2+3√2))

The secant line is
y = 1/3 + 1/9 (x-1)

The graph of f(x) and the tangent and secant lines there are shown at


http://www.wolframalpha.com/input/?i=plot+y%3Dx%2F%28x%2B2%29%2C+y%3D+%281-%E2%88%9A2%2F3%29+%2B+1%2F9+%28x-%28-2%2B3%E2%88%9A2%29%29%2C+y+%3D+1%2F3+%2B+1%2F9+%28x-1%29+for+x%3D0..5

Answer 2

Oh wow, thank you so much!