solutions_09_21

solutions_09_21 - time of day on both days? Why or why not....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 1a: Intermediate Value Theorem (Solutions) September 21, 2009 1. Is there a number x such that x 2 = cos( x )? Solution. Yes, there is. Consider the function f ( x ) = x 2 - cos( x ). We have f (0) = - 1 and f ( π/ 2) = ( π/ 2) 2 ; hence f (0) < 0 < f ( π/ 2). Since the function is continuous on the interval [0 ,π/ 2], the intermediate value theorem says that there is an x in [0 ,π/ 2] such that f ( x ) = 0, or equivalently x 2 = cos( x ). 2. Prove that there is a number which is one less than its cube. Proof. We want to show that there is an x such that x = x 3 - 1. Consider f ( x ) = x - x 3 . Then f (0) = 0 and f (2) = - 6. So, f (2) < - 1 < f (0). Since - 1 is between f (0) and f (2) and f is continuous on [0 , 2], we use the intermediate value theorem to conclude that there is an x such that f ( x ) = - 1. In other words, x - x 3 = - 1; that is x = x 3 - 1. 3. A monk leaves the monastery at 7:00 AM and takes his usual path to the top of the mountain, arriving at 7:00 PM. The following morning, he starts at 7:00 AM at the top and takes the same path back to the monastery, arriving at 7:00 PM. Is there a point on the path that the monk will cross at exactly the same
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: time of day on both days? Why or why not. Solution. This one is fun to do, so I wont give the full solution. A hint is to consider the following functions f ( x ) = The monks altitude at time x on day 1 , g ( x ) = The monks altitude at time x on day 2 Then f and g are continuous functions on the interval [7 am , 7 pm]. We want to show that f ( x ) = g ( x ) for some x in [7 am , 7 pm]. Equivalently, we want to show that f ( x )-g ( x ) = 0. Now consider the value of f ( x )-g ( x ) for x = 7 am and x = 7 pm and see what you get. 4. Prove that the polynomial x 3 + 2 x 2-5 x + 1 has at least three zeros. (Hint: consider f (-4), f (0), f (1) and f (2).) Solution. Use the hint. That is, compute f (-4), f (0), f (1) and f (2) and see what you get. 5. (Challenge problem) Show that at any given time there are two diametrically opposite points on the globe that have the same temperature. 1...
View Full Document

This note was uploaded on 03/27/2012 for the course MATH 1a taught by Professor Benedictgross during the Fall '09 term at Harvard.

Ask a homework question - tutors are online