Math 2250 Exam #1 Practice Problem Solutions. g(x) = x., h(x) =

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1 Math 50 Eam # Practice Problem Solutions. Find the vertical asymptotes (if any) of the functions g() = + 4, h() = 4. Answer: The only number not in the domain of g is = 0, so the only place where g could possibly have a vertical asymptote is at = 0. Since g will have a vertical asymptote at = 0 if either 0 + g() or 0 g() is infinite. But clearly ( + ) = +, 0 + so we know that g does indeed have a vertical asymptote at = 0. The only numbers not in the domain of h are those so that 0 = 4 = ( )( + ), so the only numbers not in the domain of h are = ±. Since and h() = = + 4 h() = =, the function h has vertical asymptotes when at both = and =.. Evaluate (a) (a) We can factor the numerator as and the denominator as Therefore, (b) + + (c) 4 = ( + )( ) = ( )( 3) = ( + )( ) ( )( 3) = + 3 = 4 = 4. (b) When <, the quantity + is negative, so Hence, + = ( + ). + + = ( + ) =. + On the other hand, when >, the quantity + is positive, so + = +. Therefore, = =.

2 Since the its from the left and right don t agree, does not eist. + + (c) Dividing numerator and denominator by 3, we get that = ( ) 3 ( ) = = 4 6 = Evaluate Answer: The numerator factors as = ( + 6)( 6), while the denominator factors as 3 6 = (3 + )( 6). Therefore, = ( + 6)( 6) )( 6) = = 0 = Evaluate Answer: The highest power of in the denominator is clearly. In the numerator, the factor is under the cube root, so we consider the leading term to be 3 = /3. Therefore, looking at the entire epression, the highest power of is just. numerator and denominator by, we see that Therefore, dividing = = = = 0. (7 9999) 3 ( )

3 5. Let f() = f is continuous provided c equals what value? { c 3 if c + if > Answer: Since both c 3 and c + are polynomials, they re continuous everywhere, meaning that f() is continuous everywhere ecept possibly at =. In order for f to be continuous at, it must be the case that f() = f(). Now, f() = ( c 3 ) = c() 3 = 4c 3, which is also the value of f(). On the other hand, f() = (c + ) = c() + = c f will be continuous when these two one-sided its are equal, meaning when Solving for c, we see that f is continuous when 4c 3 = c +. c = Is the function f defined below continuous? If not, where is it discontinuous? if < 0 f() = 3 if 0 < 3 (3 ) if 3 Answer: Since each of the three pieces of f is continuous, the only possible discontinuities of f occur where it switches from one piece to another, namely at = 0 and = 3. At = 3, and 3 f() = (3 ) = 0 3 f() = (3 ) = 0, so the right-handed and left-handed it agree. Therefore, f() = 0, 3 which is equal to f(3), so we conclude that f is continuous at = 3. On the other hand, whereas so f is discontinuous at = 0. f() = = 0, 0 0 f() = (3 ) = 3, Let f() be continuous on the closed interval [ 3, 6]. If f( 3) = and f(6) = 3, then which of the following must be true? 3

4 (a) f(0) = 0 (b) f() 3 for all between 3 and 6. (c) f(c) = for at least one c between 3 and 6. (d) f(c) = 0 for at least one c between and 3. Answer: The only one of these statements which is necessarily true is (c): since is between f( 3) = and f(6) = 3, the Intermediate Value Theorem guarantees that there is some c between 3 and 6 such that f(c) =. It s coming up with eamples to see that none of the other possibilities has to be true. 8. Find the one-sided it 4 Answer: Notice that, as, the numerator goes to, while the denominator goes to zero. Hence, we would epect the it to be infinite. However, it could be either or +, so we need to check the sign of the denominator. When <, the quantity 4 >, so 4 > 0. Therefore, in the one-sided it, the denominator is always positive. Since the numerator goes to, which is negative, the one-sided it 9. Let 4 =. f() =. f(+h) f() Does h 0 h eist? If so, what is this it? Answer: This it does not eist. To see this, I will eamine the two one-sided its: since h = h when h < 0. On the other hand, since h = h when h > 0. f( + h) f() ( + h) = h 0 = h = h = = f( + h) f() ( + h) = h 0 + h h 0 + h h = h 0 + h h = h 0 + h = Therefore, since the two one-sided its don t agree, the it does not eist. 4

5 0. Evaluate Answer: The highest power of that we see is just (since the in the numerator is under a square root). Therefore, dividing both numerator and denominator by yields = (7 + ) = = 4 7 = 7. ( ) Let f() = { 4 for a for = If f() is continuous at =, then find the value of a. Answer: In order for f to be continuous at =, we must have that a = f() = f() = 4. Now, we can factor the denominator in the it to get 4 = ( + )( ) = so we see that, in order for f to be continuous, a must be 4.. Are there any solutions to the equation cos =? Answer: Yes, there is a solution to the equation. To see this, let f() = cos. + = 4, Then f() = 0 precisely when is a solution to the equation cos =, so the problem is to show that f() = 0 for some. To see this, notice that f is continuous (since both cos and are continuous functions) and f(0) = cos 0 0 = > 0 f(π) = cos π π = π < 0. Therefore, by the Intermediate Value Theorem there eists c between 0 and π such that f(c) = 0. Then, as noted above, cos c = c, so c is a solution to the equation. Intuitively, we can see that cos = has a solution by looking at the graphs of y = cos and y = (see below); they intersect in eactly one point, so the solution c that we proved eists is actually the only solution. 5

6 3 Π Π Π Π Π 3 Π 3. Determine the following its, if they eist (a) + Answer: As the numerator goes to 4 and the denominator goes to, so (b) Answer: Notice that ++ + = 4 =. ( + + ) = + + = 4, so the numerator is going to 4. Also, the denominator is going to zero, so we epect the it to be ±. To see which, notice that, if <, then < 0, so the denominator is a very small negative number as. Hence, + + =. 4. For each of the following, either evaluate the it or eplain why it doesn t eist. (a) Answer: I m going to consider h() = 4 6. Rationalizing the denominator yields = = 6 = 6 6 ( 6)(4 + ) ( 6) ( 6)(4 + ) = = 8. 6

7 (b) Answer: Notice that the highest power of is (since the 4 is under a square root). evaluate this it, then, I want to multiply numerator and denominator by : ( ) = 6 6 =. ( ) To Since 9 4 goes to zero as, we see that 6 = Therefore, we can conclude that = 6 7. (c) Answer: I claim that this it does not eist. To see why, notice, first of all, that + ( + )( ) = ( + )( + ) = + so long as. Therefore, if it eists, must be equal to Notice that, as, the numerator goes to 3, while the denominator goes to zero. However, the sign of the denominator depends on which direction approaches from. When approaches from the left, we have that + = +, since both the numerator and denominator are negative. However, as approaches from the right, we have that + + =, since the numerator is negative and the denominator is positive. Therefore, since the two one-sided its do not agree, the given it does not eist. ( 5. (a) At which numbers is the function h() = cos ) continuous? Justify your answer. Answer: I claim that h() is continuous whenever ±. To see this, notice that g() = is a rational function, so it is continuous wherever it is defined. Since this function is defined so long as 0, we see that it is defined for all ±. In turn, the function f() = cos is continuous, and we know that the composition of continuous functions is continuous. Hence, h() = (f g)() is continuous wherever it is defined, namely for all ±. 7

8 (b) What is 0 h()? Eplain your reasoning. Answer: From part (a), we know that h() is continuous at = 0. Therefore, by definition of continuity, ( ) 0 h() = h(0) = cos 0 0 = cos(0) =. 8