# [SOLVED] Racetrack Problems

• Oct 29th 2009, 07:11 PM
iheartthemusic29
[SOLVED] Racetrack Problems
1). Consider the functions f(x)=lnx and g(x)=x−1. These are continuous and differentiable for x>0. In this problem we use the Racetrack Principle to show that one of these functions is greater than the other.

(a) Find a point c such that f(c)=g(c). c=???

(b) Find the equation of the tangent line to f(x)=lnx at x=c for the value of c that you found in (a).
For part a, I thought the answer was c=1, but that's not correct.

2). Suppose that f(t) is continuous and twice-differentiable for t>=0. Further suppose f''(t)<=9 for all t>=0 and f(0)=f'(0)=0.

Using the Racetrack Principle, what linear function g(t) can we prove is greater than than f(t) (for t>=0)?
g(t)= ???

Then, also using the Racetrack Principle, what quadratic function h(t) can we prove is greater than than f(t) (for t>=0)?
h(t)= ???

I have NO clue!
• Oct 29th 2009, 07:31 PM
ux0
Quote:

Originally Posted by iheartthemusic29
1). Consider the functions f(x)=lnx and g(x)=x−1. These are continuous and differentiable for x>0. In this problem we use the Racetrack Principle to show that one of these functions is greater than the other.

(a) Find a point c such that f(c)=g(c). c=???

(b) Find the equation of the tangent line to f(x)=lnx at x=c for the value of c that you found in (a).
For part a, I thought the answer was c=1, but that's not correct.

Why is c=1 not correct?

a)
$\displaystyle f(1)=ln(1)=0$
$\displaystyle g(1)=1-1=0$

b) the slop of the tangent line is just $\displaystyle f'(x)= \frac{1}{x}$ ... use that to figure out the equation of the line..
• Oct 29th 2009, 08:57 PM
iheartthemusic29
I think there's an error in the answer key, because, as you said, c should equal 1.

Can someone help me with question 2??? I don't even know where to begin on this one. Please! And thanks!
• Oct 30th 2009, 12:42 PM
iheartthemusic29
Never mind! I figured it out. (Sun)
g(t)=9t and h(t)=(9/2)t^2