f10finalsolutions

f10finalsolutions - Fall 2010 Math 113 Final Exam Solutions...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Fall 2010 Math 113 Final Exam Solutions 1. If f ( x ) = x ln x , find f ( e 3 ). Answer: Using the quotient rule, f ( x ) = ln x d dx ( x )- x d dx (ln x ) (ln x ) 2 = ln x 1- x 1 x (ln x ) 2 = ln x- 1 (ln x ) 2 . Therefore, f ( e 3 ) = ln( e 3 )- 1 (ln( e 3 )) 2 = 3- 1 3 2 = 2 9 . 2. At what value(s) of x (if any) is the function f ( x ) defined below discontinuous? f ( x ) = x 2 + 4 x + 5 if x <- 2 x 2 if- 2 x 2 1 + x- 2 if x > 2 For x <- 2, f ( x ) = x 2 + 4 x + 5, which is a polynomial and, hence, continuous. Also, for- 2 < x < 2, f ( x ) = x 2 , which is also a polynomial and hence continuous. Finally, for x > 2, f ( x ) = 1 + x- 2, which is continuous since its the composition of continuous functions (specifically, f ( x ) = g h ( x ) where g ( u ) = 1 + u and h ( x ) = x- 2). Therefore, the only possible points of discontinuity for f are at x =- 2 and x = 2. At x =- 2, f ( x ) will be continuous if lim x - 2 f ( x ) = f (- 2) =- 2 2 =- 1 . Clearly, lim x - 2 + f ( x ) = lim x - 2 + x 2 =- 1. On the other hand, lim x - 2- f ( x ) = lim x - 2- ( x 2 + 4 x + 5 ) = (- 2) 2 + 4(- 2) + 5 = 4- 8 + 5 = 1 . Therefore, since lim x - 2 + f ( x ) =- 1 and lim x - 2- f ( x ) = 1, we can see that f ( x ) is not continuous at x =- 2. Turning to x = 2, f ( x ) will be continuous if lim x 2 f ( x ) = f (2) = 2 2 = 1 . Now lim x 2- f ( x ) = lim x 2- x 2 = 1. On the other hand lim x 2 + f ( x ) = lim x 2 + ( 1 + x- 2 ) = 1 . Since the limits from the left and the right are both equal to 1, we conclude that lim x 2 f ( x ) = 1 , so f ( x ) is continuous at x = 2. Therefore, f ( x ) is continuous at all real numbers except x =- 2. 1 3. Find the equation of the tangent line to the curve x cos y = 1 at the point ( 2 , 3 ) . Answer: We will find the slope of the tangent line using implicit differentiation. Differentiating both sides yields 1 cos y + x (- sin y y ) = 0 or cos y- xy sin y = 0 . Therefore, cos y = xy sin y and so we have that y = cos y x sin y ....
View Full Document

Page1 / 7

f10finalsolutions - Fall 2010 Math 113 Final Exam Solutions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online