31B_pracmid1 - MATH 31B Lecture 4 Fall 2006 Practice Midterm 1 NAME STUDENT ID This is a closed-book and closed-note examination Calculators are not

# 31B_pracmid1 - MATH 31B Lecture 4 Fall 2006 Practice...

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MATH 31B - Lecture 4 - Fall 2006 Practice Midterm 1 - October 25, 2006 NAME: STUDENT ID #: This is a closed-book and closed-note 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 ) 0 (3). Solution. Note that f (0) = 3 and f 0 ( x ) = 2 x + π 2 sec 2 πx 2 . Thus ( f - 1 ) 0 (3) = 1 f 0 (0) = 1 π/ 2 = 2 . 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 = ln 3 or x = ln 4.
3 3. Find the derivative of the function f ( x ) = (ln x ) x . Solution. If y = f ( x ) then ln y = x ln ln x . Taking the derivative implic- itly, y 0 /y = x 1 x ln x + ln ln x and hence f 0 ( x ) = y 1 ln x + ln ln x = (ln x ) x 1 ln x + ln ln x . 4. Evaluate the limit lim x 1 1 - x + ln x 1 + cos( πx ) . Solution. Since lim x 1 (1 - x + ln x ) = 0 and lim x 1 (1 + cos( πx )) = 0 L’Hospital’s rule applies. Thus lim x 1 1 - x + ln x 1 + cos( πx )
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