This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: MATH 31B  Lecture 4  Fall 2006 Practice Midterm 1  October 25, 2006 NAME: STUDENT ID #: This is a closedbook and closednote examination. Calculators are not allowed. Please show all your work. Use only the paper provided. You may write on the back if you need more space, but clearly indicate this on the front. 1 2 1. Let f ( x ) = 3 + x 2 + tan( πx 2 ) , 1 < x < 1. Find the value of the derivative ( f 1 ) (3). Solution. Note that f (0) = 3 and f ( x ) = 2 x + π 2 sec 2 πx 2 . Thus ( f 1 ) (3) = 1 f (0) = 1 π/ 2 = 2 /π . 2. Solve for x : 7 e x e 2 x = 12 Solution. If z = e x then the equation becomes z 2 7 z + 12 = 0, i.e. ( z 3)( z 4) = 0. Thus either e x = 3 or e x = 4 and hence either x = ln3 or x = ln4. 3 3. Find the derivative of the function f ( x ) = (ln x ) x . Solution. If y = f ( x ) then ln y = x lnln x . Taking the derivative implic itly, y /y = x 1 x ln x + lnln x and hence f ( x ) = y 1 ln x + lnln x ¶ = (ln x ) x 1 ln x + lnln x ¶ ....
View
Full Document
 Fall '08
 VALDIMARSSON
 Math, Calculus, lim, ln ln, ln x ln

Click to edit the document details