Solution to Review Problems for Midterm #1

Transcription

1 Solution to Review Problems for Midterm # Midterm I: Wednesday, September in class Topics:.,.3 and.-.6 (ecept.3) Office hours before the eam: Monday - and 4-6 p.m., Tuesday - pm and 4-6 pm at UH 080B) Topics Find the equation of the tangent line to y f() at a,. We need to find the slope of the tangent line by finding the limit f(a+h) f(a) m lim h 0 h. Use the point slope formula to find the equation of the tangent line thru (a, f(a)) with slope m: y f(a) m( a).. Let f() Find an equation of the tangent line to the curve at P (, f()). Solution: To find the slope of the tangent line at, we need to f(+h) f() find the limit m lim h 0. Since f( ) + 3 +, we h have f( + h) ( + h) + 3( + h) + }{{} ( + h + h ) + } 3 + 3h {{ + } 4h h + } 4 + {{ 3h } h h, f() and f( + h) f() h h h h h( h). So m f(+h) f() lim h 0 h h( h) lim h 0 lim h h 0 h. Now we have the point P (, f()) (, ) and the slope of the tangent line m. By the point slope formula, we have the equation of the tangent line y ( ) + and y Hence the equation of the tangent line to the curve at P (, f()) is y Let f(). Find an equation of the tangent line to the curve at + P (, f()). Solution: To find the slope of the tangent line at, we need f(+h) f() to find the limit m lim h 0. Since f( ), we have h + f( + h), f() and (+h) + (+h+h )+ +h+h + f(+h) f() +h+h (+h+h ) h h +h+h +h+h +h+h +h+h +h+h h h +h+h lim h 0 h( h) f(+h) f() +h+h. So m lim +h+h h 0 lim h( h) h h 0 lim h h 0. Now we have the point P (, f()) h( h) ( h) (+h+h ) (+h+h )h (, ) and the slope of the tangent line m. By the point slope formula, we have the equation of the tangent line y () + and y Hence the equation of the tangent line to the curve at P (, f()) is y +. MATH 850: page of 8

2 Solution to Review Problems for Midterm # MATH 850: page of 8 3. Let f(). Find an equation of the tangent line to the curve at +3 P (, f()). Solution: To find the slope of the tangent line at, we need f(+h) f() to find the limit m lim h 0. Since f( ) h, we have +3 f( + h) (+h) +3 (+h+h )+3, f() and +3 f( + h) f() rationalize the epression by multiplying + + to get f( + h) f() h h ) ( )(+ ) h h ) ( )(+ ). f(+h) f() So m lim h 0 h h( h)) lim h 0 ( )(+ )h ( ) ( )(+ ) h h ) ( )(+. Now we 4+h+h lim ) h 0 h (factoring out h from the top and the ( h) ( )(+ ) 0 4 (+ 4) bottom) lim h 0. Now we 4 4 have the point P (, f()) (, ) and the slope of the tangent line m. By the point slope formula, we have the equation of the 4 tangent line y 4 ( ) and y Hence the equation of the tangent line to the curve at P (, f()) is y Topic Left continuity: To determine if a function is left continuous at a, we need to find lim a f() and f(a). If lim a f() f(a) then f is not left continuous. If lim a f() f(a) then f is left continuous. Right continuity at a: To determine if a function is right continuous at a, we need to find lim a +f() and f(a). If lim a +f() f(a) then f is not left continuous. If lim a +f() f(a) then f is left continuous. Continuity at a:to determine if a function is t continuous at a, we need to find lim a f(), lim a +f() and f(a). If lim a f() lim a +f() then f is not continuous. (either jump or infinite discontinuity). If lim a f() lim a +f() f(a) then f is not continuous. (This is called the removable discontinuity.) If lim a f() lim a +f() f(a) then f is continuous.

3 MATH 850: page 3 of 8 Solution to Review Problems for Midterm # 4. A piecewise defined function is given by, < f(), < +, Determine if f is left continuous, right continuous or continuous at or. Solution: At, we have lim f() lim 0, lim + f() lim + ( ) 0 and f() 0. So lim f() lim + f() f(). Hence f is left continuous, right continuous and continuous at. At, we have lim f() lim 4 3, lim + f() lim and f() + 4. So lim f() f() and f is not left continuous at. Now lim + f() f() and f is right continuous. But lim f() 3 4 lim + f() and f has a jump discontinuity at. 5. Classify the discontinuity of the following functions (removable, infinite, jump or oscillating discontinuity). Redefine the value of the function if it s removable. (a) f() Solution: Note that ( )( 3). The domain of f is { and 3}. So f is discontinuous at and 3. At 3, lim lim lim ()( 3) () 0 3 and lim + lim So f has an infinite () 0 + discontinuity at. 3 At 3, lim 3 lim and lim 3 () lim 3 +. So f has an removable discontinuity at 3. () We can define f(3) to make f continuous at 3. (b) f() +, < +, 8, <

4 Solution to Review Problems for Midterm # MATH 850: page 4 of 8 Solution: The only points where f may not be continuous is and. At, lim f() lim + lim + lim (+)(), lim + f() lim and f( ) + 0. Hence 8 8 lim f() f( ) and f is not left continuous at Now lim + f() f( ) and f is right continuous at. Since lim f() 0 lim + f() and f has a jump discontinuity at. At, lim f() lim +, lim f() lim + lim ( )( +) + ( )( +) lim + ( )( lim +) + ()(+)( lim +) + and f() +. Since lim 8 4 f() lim + f() f() and f has is left continuous, right continuous and continuous at. (+)( +) 4 6. A piecewise defined function is given by, < f() a + b, <, (a) Find the graph of y f() over the interval (, ) [, ). (b) Determine the value of a and b so that f is continuous everywhere. Also eplain your answer geometrically. Solution: The only possible discontinuity are at and. We first compute lim f() lim, lim + f() lim + a + b a + b and f( ) a + b To make f continuous at, we must have lim f() lim + f() f( ). This gives a + b. Net we compute lim f() lim a+b a+b, lim + f() lim + and f() To make f continuous at, we must have lim f() lim + f() f(). This gives a + b. Now we have two equations a + b and a + b. From a + b, we get b + a. Plugging b + a to a + b, we get a + ( + a), a 3 and a 3. Use b + a to get b + 3. Thus a 3 and b will make f continuous everywhere.

5 MATH 850: page 5 of 8 Solution to Review Problems for Midterm # Topic 3: Vertical asymptote and horizontal asymptote. To find vertical asymptote, we first find its domain. Let a be a point not in the domain. We find lim a f() and lim a f(). If lim a f(), lim a f(), lim a + f() or lim a + f() then a is a vertical asymptote. To find horizontal asymptote, we find lim f() and lim f(). If lim f() b then y b is a horizontal asymptote. If lim f() c then then y c is a horizontal asymptote. 0 + negative number 0, 0 positive number 0, , ,,, ( )odd power. p n n+p, lim positive number, lim negative number 0, lim positive and even integer, lim positive and odd integer. Given a rational function R(). Suppose we rearrange the order of the power in the top and the bottom so it s in decreasing order, i.e R() ann +a n n + +a +a 0 b m m +b m m + +b +b 0. For eample, R() a lim n n +a n n + +a +a 0 ± b m m +b m m + +b +b 0 lim n (a n+a n + +a n+ +a 0 n ) ± a lim n b m n m. 7. Determine the following limits (a) lim 3+ Solution:lim 3+ lim m (b m+b m + +b m+ +b 0 m ) ()( ) lim. 4 (b) lim +, lim 4 5+6, lim Solution: Plugging in the epression, we get. So we need 0 to factor the bottom to analyze its behavior. lim + 4 lim +. ( )( 3) lim lim ( )( 3) Now we know that lim + lim So lim 5+6 doesn t eist (c) lim + 6 +, lim Solution: lim lim 6 ( ) lim 3 ( ) 3 ( ) ( ) lim 3 (. Note that we have used the fact that lim ) ( ). Similarly, lim lim 6 ( ) lim 3 ( ) 3 ( ) ( ) lim 3 ( ) 3.

6 Solution to Review Problems for Midterm # MATH 850: page 6 of 8 (d) lim 6 + +, lim Solution: lim lim 6 ( ) ( ) (factoring out 6 ( + from the top and the bottom) lim ) ( ). 6 3 Similarly, lim (e) lim 6 + +, lim Solution: lim ( 5) Similarly, lim lim 0. ( 5) lim 6 ( ) lim 6 ( ) ( ) ( ) lim ( ) lim 8 ( ) ( ) ( ) lim ( ) lim 8 ( ) ( ) (f) lim Solution: lim 4 + lim + ( +) 4 + lim ( 4 +) + ( lim ). (+ ) lim ( ) lim lim lim + ( ) (+ ) (g) lim + Solution: lim + lim ( + ) lim ( +) ( ) ++ lim lim ++ lim lim lim ( + + ) (h) lim + Solution:lim + lim ( + ) lim ( +) () ++ lim lim + ++ lim ( + ) ( + + ) lim ( + ) ( + + ) lim lim ( + ) ( + + ). 8. Find the domain of the following functions and determine the vertical and horizontal asymptotes of the graph of the following functions.

7 MATH 850: page 7 of 8 Solution to Review Problems for Midterm # (a) f() 3+ Solution: To find the domain, we factor 3 + ( )( ). So the domain is { and }. At, lim 3+ lim lim ()( ) and lim ( ) + lim 3+ + lim +. So lim ( ). Hence is not a 3+ vertical asymptote. In fact, has a removable discontinuity 3+ at. At, lim lim 3+ lim ()( ) ( ) 0 and lim + lim 3+ + Hence is a vertical asymptote. lim ( ) lim ()( ) +. ( ) 0 + ( ) ( 3 + ) lim 3+ lim ( 3 + ) lim lim ( ) 0 0. Similarly, lim ( 3 + ) ( lim ) ( 3 + ) lim 0 0. Hence y 0 is a hori- lim lim zontal asymptote. ( ) ( 3 + ) ( ) ( 3 + ) 3+ (b) f() 5+6 To find the domain, we factor ( )( 3). the domain is { and 3}. At, lim lim ( )( 3) 0 ( ) and lim lim ( ) Hence is a vertical asymptote. 0 At 3, lim 3 lim lim 3 + lim a vertical asymptote. lim ()( ) So 5+6 ( )( 3) and ( )( 3) 0 0. Hence 3 is ( )( 3) ( ) ( ) ( 5 +6 ) lim 5+6 lim ( 5 +6 ) lim lim ( ) 0 0. Similarly, lim ( 5 +6 ) lim lim ( ) 0 0. Hence y 0 is a horizontal ( 5 +6 ) asymptote. (c) f() To find the domain, we factor ( )( 3). the domain is { and 3}. At, lim 3 lim ( ) ( )( 3) 0 ( ) 7 and lim At 3, lim 3 Hence is a vertical asymptote. 0 3 lim ( )( 3) 5+6 So lim and 3 ( )( 3)

8 Solution to Review Problems for Midterm # MATH 850: page 8 of 8 lim lim a vertical asymptote. 3 ( )( 3) Hence 3 is lim 3 lim 3 ( 3 ) 5+6 lim ( 5 +6 ) ( 3 ) ( 5 +6 ) ( lim lim ). Similarly, lim ( 5 +6 ) ( lim lim ) ( 5 +6 ) a horizontal asymptote.. Hence y 3 ( )( 3) doesn t have (d) f() To find the domain, we factor ( )( 3). the domain is { and 3}. At, lim + lim + ( )( 3) ( ) ( ) So 5+6 and lim lim Hence is a vertical asymptote. 0 At 3, lim 3 + lim ( )( 3) lim lim ( )( 3) is a vertical asymptote. 8 8 and Hence lim + lim 5+6 ( + ) lim ( 5 +6 ) ( + ) ( 5 +6 ). Similarly, lim + lim 5+6 ( + ). Hence ( 5 +6 ) y is a horizontal asymptote. ( )( 3)