Exam 1 Solutions. x(x 2) (x + 1)(x 2) = x

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1 Eam Solutions Question (0%) Consider f() = (a) By calculating relevant its, determine te equations of all vertical asymptotes of te grap of f(). If tere are none, say so. f() = ( 2) ( + )( 2) = ( + ) + = 0 = = is a vertical asymptote (2) Verify: f() = () = (3) (b) By calculating relevant its, determine te equations of all orizontal asymptotes of te grap of f(). If tere are none, say so. f() + + = () f() = + = + = (2) Terefore, y = is te only orizontal asymptote.

2 2 Question 2 (2%) Calculate te it, if it eists; if it does not eist, say so. Justify your answers, but do not use evidence from numerical approimations. (a) ( )( ) (b) Since terefore And so by te Squeeze Teorem ( ) = 2 3 sin > 0 for > 0 and sin () sin. (2) ( ) = 0 (3) sin = 0 (4)

3 3 Question 3 (2%) Calculate te it, if it eists; if it does not eist, say so. Justify your answers, but do not use evidence from numerical approimations. (a) (b) (6 t 3 )(t 3) t 2 (6 t 3 )(t 3) = (6 2 3 )(2 3) = 2 t ( 2)( + ) = 3

4 4 Question 4 Let f() = + 2. (a) Find te domain [a, b] of te function f(). Te domain of + is [, ), and te domain on 2 is (, 2], so te domain of f() is [, 2], te intersection of tese two intervals. (b) f is a root function, terefore, f() is continuous in its domain. Use te Intermediate Value Teorem to sow tat te equation f() = as a solution in (a, b). Since f ( ) = 3 < f (2) = 3 >. Ten by te Intermediate Value Teorem, tere eists some c (, 2) suc tat f(c) =.

5 5 Question 5 (2%)Find te derivative of te function f() = using te definition of te derivative. Do NOT use any rules of differentiation tat you may ave learned in a previous calculus class. f () 0 f( + ) f() ( + ) 0 0 ( + ( + ) = 2 ( + ) ) + ) ( + ) 0 ( + )

6 6 Question 6 (2%) For te function f() = 2 3, we know tat 2 f() =. (a) Fill in te blanks below to give a precise definition of te it: For every ε > 0 tere eists some δ > 0 suc tat if, ten. For every ε > 0 tere eists some δ > 0 suc tat if 0 < 2 < δ ten f() < ϵ (b) Using your answer to (a), find a δ corresponding to ε. graping te function! () Analysis: Do tis algebraically, not by 2 3 < ϵ = 2 4 < ϵ = 2( ) < ϵ = < ϵ 2. Take δ = ϵ 2 (2) Proof: For any ϵ > 0, take δ = ϵ 2. Wenever 2 < δ = ϵ 2 = 2 2 < ϵ = 2 4 < ϵ = 2 3 < ϵ = f() < ϵ

7 7 Question 7(2%) Let f() = { k + 2 if 2 if >. (a) Find a value of k tat makes f continuous at =. You must sow tis algebraically, witout using a table or grap. f will be continuous at = if f() + f() Setting k + 2 = gives us k = f() k + 2 = k + 2 f() 2 = + + (b) For te value of k you found in (a), is f differentiable at =? Wy or wy not? f( + ) f() 0 0 f( + ) f() = ( + ) Tese its are unequal, so f is not differentiable at = = 2

8 8 Question 8 (2%) Consider te curve y = 2. (a) Find te slope of te tangent line to te curve at te point (, ) by using te definition of te derivative at a point. Let f() = 2 f f( + ) f() () 0 ( + ) 2 ) (b) Find an equation of te tangent line to te curve y = 2 troug te point (, ). y = m( ) = 2 2, so te tangent line is y = = 2 MAKE-UP Eam Solutions Question (0%) Consider (a) By calculating relevant its, determine te equations of all vertical asymptotes of te grap of f(). If tere are none, say so. f() = ( + ) ( + )( 4) = ( 4) 4 = 0 = = 4 is a vertical asymptote (2) Verify: f() = () 4 = (3) (b) By calculating relevant its, determine te equations of all orizontal asymptotes of te grap of f(). If tere are none, say so. f() 4 4 = () f() = 4 = 4 = (2) Terefore, y = is te only orizontal asymptote.

9 9 Question 2 (2%) Calculate te it, if it eists; if it does not eist, say so. Justify your answers, but do not use evidence from numerical approimations. (a) (b) Since terefore And ( ) 0 so by te Squeeze Teorem ( )( ) ( )( ) sin = 0 9 ( ) 2 > 0 for > 0 and sin( ) () sin( ). (2) = 0 (3) sin( 0 + ) = 0 (4)

10 0 Question 3 (2%) Calculate te it, if it eists; if it does not eist, say so. Justify your answers, but do not use evidence from numerical approimations. (a) (b) (2 t + t 2 )(5t 3 t 2 + ) t (2 t + t 2 )(5t 3 t 2 + ) = (2 + 2 )( ) = 0 t ( 2) ( 2)( + 2) =

11 Question 4 Let f() = (a) Find te domain [a, b] of te function f(). Te domain of 2 + is [ 2, ), and te domain on 5 is (, 5], so te domain of f() is [ 2, 5], te intersection of tese two intervals. (b) f is a root function, terefore, f() is continuous in its domain. Use te Intermediate Value Teorem to sow tat te equation f() = as a solution in (a, b). Since f ( 2) = 7 < f (5) = 7 >. Ten by te Intermediate Value Teorem, tere eists some c ( 2, 5) suc tat f(c) =.

12 2 Question 5 (2%)Find te derivative of te function f() = using te definition of te derivative. Do NOT use any rules of differentiation tat you may ave learned in a previous calculus class. f () 0 f( + ) f() ( + ) 2 + 2( + ) + ( 2 2 ) = 2 + 2

13 3 Question 6 (2%) For te function f() = 5 4, we know tat f() =. (a) Fill in te blanks below to give a precise definition of te it: For every ε > 0 tere eists some δ > 0 suc tat if, ten. For every ε > 0 tere eists some δ > 0 suc tat if 0 < < δ ten f() < ϵ (b) Using your answer to (a), find a δ corresponding to ε. graping te function! () Analysis: Do tis algebraically, not by 5 4 < ϵ = 4 4 < ϵ = 4( ) < ϵ = < ϵ 4. Take δ = ϵ 4 (2) Proof: For any ϵ > 0, take δ = ϵ 4. Wenever < δ = ϵ 4 = 4 < ϵ = 4 4 < ϵ = 4 4 < ϵ = 5 4 < ϵ = f() < ϵ

14 4 Question 7 Let f() = { k + 5 if 2 + if >. (a) Find a value of k tat makes f continuous at =. You must sow tis algebraically, witout using a table or grap. f will be continuous at = if f() + f() f() k + 5 = k + 5 f() 2 + = Setting k + 5 = 2 gives us k = 3 (b) For te value of k you found in (a), is f differentiable at =? Wy or wy not? f( + ) f() f( + ) f() = ( + ) Tese its are unequal, so f is not differentiable at = = 2

15 5 Question 8 (2%) Consider te curve y = 2 +. (a) Find te slope of te tangent line to te curve at te point (, 2) by using te definition of te derivative at a point. Let f() = 2 + f f( + ) f() () 0 ( + ) 2 + 2) (b) Find an equation of te tangent line to te curve y = 2 troug te point (, 2). y 2 = m( ) = 2 2, so te tangent line is y = =