Math 5 Spring Written Homework 0 Solutions. For following its, state what indeterminate form the its are in and evaluate the its. (a) 3x 4x 4 x x 8 Solution: This is in indeterminate form 0. Algebraically, we hope to be able to factor the 0 numerator and denominator and convert the argument of the it to a function defined at x =. 3x 4x 4 x x 8 (3x + )(x ) x (x + )(x ) 3x + since, by properties of its, x x (x + ) 3() + ( + ) = 8 8 = (b) t 8t 3 + t (t )(t + ) Solution: This is the it of a rational function. The associated indeterminate form is. t 8t 3 + t (t )(t + ) t 8t 3 + t 4t 3 t + t 8t 3 + t t 4t 3 t + t /t3 /t 3 8 + /t t 4 /t + /t /t 3 8 + 0 = 4 0 + 0 0 =
(c) ( + h) 3 h 0 h Solution: This is indeterminate form 0 0. ( + h) 3 h 0 h ( + 3h + 3h + h 3 ) h 0 h 3h + 3h + h 3 h 0 h (3 + 3h + h ) h 0 = 3 + 0 + 0 = 3 (d) h 0 4 + h h Solution: This is indeterminate form 0 0. h 0 4 + h h h 0 4 + h h (4 + h) 4 h 0 h h h 0 h 4 + h + h 0 4 + h + = 4 + 0 + ( ) 4 + h + 4 + h + = 4 (e) (x x 4 x ) Solution: We talked of its in the form while discussing the long-run behavior of polynomials. We will re-address this as yet another indeterminate form. As with all indeterminate forms, we try an algebraic solution before attempting a sequential one. Here, as in (d), we multiply by the conjugate of x x 4 x.
(x x 4 x ) (x ( x + ) x x 4 x ) 4 x x + x 4 x (x x4 x )(x + x 4 x ) x + x 4 x x 4 (x 4 x ) x + x 4 x x + x + x 4 x x + x + x 4 x + /x + x x4 x ( ) x x + (x x 4 x ) 4 + /x + /x + /x /x 4 + 0 = + 0 0 =
. Explain why the following are not in indeterminate form. Evaluate the it. Fully justify your answer. (a) x 0 x 4 Solution: This it is in the form. We need to use sequences. 0 Notice that the x 4 term in the denominator will turn any input value x into a positive number. As a result, the it from both sides will be the same. Rather than using a it from both sides, use the sequence generated by f(n) = ( ) n /n. We know that the it of this sequence is 0, but alternates sign with each term. x 0 x 4 (( ) n /n) 4 (( ) 4n /n 4 ) (/n 4 ) n 4 = (b) ( x x 4 x 3 ) Solution: As x goes to infinity, x also goes to. For large values of x, x 4 x 3 > 0. Hence, as x goes to infinity, x 4 x 3 also goes to. Hence, ( x x4 x 3 ) =.
3. Using its, determine the horizontal asymptotes of the graphs of the following functions, if any. If the graph does not have any horizontal asymptotes, you must clearly show why. (a) f(x) := 3x 4x 4 x 8 3x 4x 4 Solution: Evaluate. Note that this it is in indeterminate form x 8. Since the degree of the polynomial in the denominator is, we multiply the quotient by /x. /x 3x 4x 4 x 8 3x 4x 4 x 8 ( ) /x /x 3 4/x 4/x 8/x = 3 0 0 since /x and /x 0 as x 0 = 3/ A horizontal asymptote occurs at y = 3 p(x). Recall that for rational functions, if q(x) p(x) is finite, approaches the same it. x q(x) (b) g(x) := 3x3 4x 4 x 8 3x 3 4x 4 Solution: Evaluate. Note that this it is in indeterminate form x 8. Since the degree of the polynomial in the denominator is, we multiply the quotient by /x. /x 3x 3 4x 4 x 8 3x 3 4x 4 x 8 3x 4/x 4/x 8/x By the same arithmetic, 3x 3 4x 4 x 8 ( ) /x /x 3x since /x and /x 0 as x = since 3x as x 3x =. The graph of this function has no horizontal asymptote.
(c) h(x) := 3x3 4x 4 x 5 8 3x 3 4x 4 Solution: Evaluate. Note that this it is in indeterminate form x 5 8. Since the degree of the polynomial in the denominator is 5, we multiply the quotient by /x5. /x 5 3x 3 4x 4 x 5 8 3x 3 4x 4 x 5 8 A horizontal asymptote occurs at y = 0. ( ) /x 5 /x 5 3/x 4/x 4 4/x 5 8/x 5 = 0 0 0 since for p > 0, /x p 0 as x 0 = 0
4. (a) How many horizontal asymptotes can the graph of a rational function have? Explain. Solution: Given a rational function R(x) := p(x), we know from lecture it will have a q(x) horizontal asymptote if the degree of q is greater than or equal to the degree of p. When degree of q is greater than p, the it to both + and of R(x) is 0. There is only one horizontal asymptote. When the degree of q is equal to the degree of p, the it is dependent on the leading coefficents of the two polynomials. If the leading term of p(x) is c n x n and the leading term of q(x) is d n x n, R(x) = x ± c n x n x ± d n x = n Again, there is only one horizontal asymptote. c n = c n. x ± d n d n If a rational function has a horizontal asymptote, the same asymptote is approached as x goes to both + and. (b) How many horizontal asymptotes can the graph of an arbitrary function have? Can a function intersect a horizontal asymptote? Explain. Solution: In general, there is no reason that f(x) and f(x) have to both exist and x be the same. The graph of an arbitrary function can have 0, one or (at most) two horizontal asymptotes. A function can intersect its horizontal asymptote. For example, the graph of the last example in lecture on rational functions did this. Horizontal asymptotes are information about what happens in the long run: as x and as x. The say nothing about what happens in the middle of the graph.
5. Evaluate the following its. (a) x 0 x + x Solution: The it is of the form 6. As a result, we will need to use x 0 x 0 + x sequences and check the it from both sides.. from the right of 0 Use the sequence O + (n) :=. By the continuity of the rational function g(x) := n, we expect g(x) x + x x 0 + g(o+ (n)). x 0 + x + x (/n) + (/n) 6 (/n) + (/n) + 6 n n n + n + 6 n n n + n n + 6 n n + n n 6 n =. from the left of 0 Use the sequence O (n) :=. By the continuity of the rational function g(x) := n, we expect g(x) x + x x 0 g(o (n)). x 0 + x + x ( /n) + ( /n) 6 ( /n) + ( /n) 6 n n n n 6 n n n n n 6 n n n n 6 n =
Since the it from the right of zero and the it from the left of zero don t grow in the same direction, we CAN NOT write = something. Hence, we say x 0 x + x does not exist. x 0 x + x x 4 6 (b) x 0 x 4 4x x 4 6 6 Solution: The it is of the form. As a result, we will need to use x 0 x 4 4x 0 sequences and check the it from both sides.. from the right of 0 Use the sequence O + (n) :=. By the continuity of the rational function r(x) := n x 4 6, we expect r(x) x 4 4x x 0 + r(o+ (n)). x 4 6 (/n) 4 6 x 0 + x 4 4x (/n) 4 4(/n) 6 n 4 n 4 4 n 6 n 4 n 4 4 n n 6n n 4 6n 4 = n n. from the left of 0 Use the sequence O (n) :=. By the continuity of the rational function r(x) := n
x 4 6, we expect r(x) x 4 4x x 0 r(o (n)). x 4 6 ( /n) 4 6 x 0 x 4 4x ( /n) 4 4( /n) 6 n 4 n 4 4 n 6 n 4 n 4 4 n n 6n n 4 6n 4 = Since the it from the right of zero and the it from the left of zero grow in the same direction, we write (c) x 4 6 x x 4 4x Solution: The it x x 4 6 x 4 4x x 0 x + x =. n n 6 is of the form. As a result, we hope an algebraic solution will suffice to answer this question. Begin by factoring r(x) := x4 6 x 4 4x. Hence, x r(x) := x4 6 x 4 4x x 4 6 (x + 4)(x 4) x 4 4x x x (x 4) = (x + 4)(x 4) x (x 4) = x + 4 x when x is not or 0 (x + 4) x x since, by properties of its, x infinitesimally close to, but not equal to = (() + 4) () since ˆr(x) := (x + 4) x continuous at x = =
(d) (x + )(x ) x x 4 Solution: Note that this is the long-run behavior of the rational function d(x) := x 3 3x + 4. The indeterminate form here is x 4. (x + )(x ) x x 4 x 3 3x + 4 x x 4 x x 3 3x + 4 x 4 x 3 + 4 x x 4 x x = x x x (x + )(x ) x 4 = 6. For each rational function given, use the end behavior, intercepts, asymptotes, and continuity to draw a rough sketch of the graph of the function. (a) g(x) := x + x 6 x + x Solution:. domain Required that x + x 0. Hence, x 0 or x. The domain of g is (, ) (, 0) (0, ).. end behavior Need to examine the following its: x, x, x, and x 0. We have already done many of these. (a) In lecture, we have shown that g(x) = 7. This tells us that there will be a x hole in the graph at (, 7/). (b) In Prob 5a, we showed = and x 0 + x + x x 0 x + x Thus, the graph has a vertical asymptote at x = 0. =.
(c) For the it to infinity, x + x /x x + x /x + 6 x x + x = Hence there is a horizontal asymptote at y =. (Recall rational functions that have a horizontal asymptote have the same long-run behavior at both ends: Prob 4.) 3. intercepts (a) y-intercept Since g(0) is undefined, there is no y intercept. (b) x-intercept(s) 4. graph We need the roots of the numerator that are in the domain of g. Since = (x 3)(x + ) = 0 has the solutions x = and x = 3/, the potential intercepts occur at x = and x = 3/. But, x = is not in the domain of g. Hence, the only x-intercept is at (3/, 0). We begin by graphing the hole, the x-intercept, the behavior as x approaches the vertical asymptote at x = 0, and the horizontal asymptote at y =. Then we use continuity to complete the graph.
(b) r(x) := x4 6 x 4 4x Solution:. domain Required that x 4 4x 0. Since x 4 4x = x (x + )(x ), this requires x 0, x or x. The domain of g is (, ) (, 0) (0, ) (, ).. end behavior Need to examine the following its: x, x, x, x, and x 0. Again, we have already done many of these. x 4 6 (a) In Prob 5b, we showed =. Thus, the graph has a vertical asymptote at x = x 0 x 4 4x 0. x 4 6 (b) In Prob 5c, we showed =. This tells us that there will be a hole x x 4 4x in the graph at (, ). (c) We have not yet checked the x x 4 6 x 4 4x case. That said, the algebra in this computation is identical to our solution to Prob 5c above. Since x x 4 6 x 4 4x is
in indeterminate for 0 0 and x4 6 x 4 4x = x + 4 x x x 4 6 x 4 4x x + 4 =. x x on r s domain, we get This tells us that there is also a hole in the graph at (, ). (d) For the it to infinity, x 4 6 x 4 6 x 4 4x x 4 4x /x4 /x 4 6/x 4 4/x = Hence there is a horizontal asymptote at y =. (Recall rational functions that have a horizontal asymptote have the same long-run behavior at both ends: Prob 4.) 3. intercepts (a) y-intercept Since r(0) is undefined, there is no y intercept. (b) x-intercept(s) We need the roots of the numerator x 4 6 that are in the domain of r. Since x 4 6 = (x + 4)(x )(x + ) = 0 has the solutions x = and x =, the potential intercepts occur at x = and x =. But, neither of these values are in the domain of r. Hence, there are no x-intercepts either. 4. graph We begin by graphing the holes, the horizontal asymptote at y =, and the behavior as x approaches the vertical asymptote at x = 0. Then we use continuity to complete the graph.
(c) p(x) := x3 3x + x + Solution:. domain Required that x + 0. The domain of g is (, ) (, ).. end behavior Need to examine the following its: x, x, and x. x 3 3x + (a) The case. Note that this it is in the indeterminate form x x + 0. As a result, we factor the numerator knowing that x = is a root of the 0 polynomial x 3 3x +. x x + Then, x 3 3x + x x + x + ) x 3 3x + x 3 x x 3x x + 4x (x + )(x x + ) x x + x + x 0 x (x x+) = ( ) ( )+ = 9.
This tells us that there will be a hole in the graph at (, 9). (b) For the it to infinity, x 3 3x + x + x 3 3x + /x x + /x x 3 + /x + / x = Hence there no horizontal asymptote for the graph of this rational function. Now, when there is no horizontal asymptote, we are not guaranteed that the end behavior will be the same on both sides. We need to check the it to. x 3 3x + x x + x 3 3x + /x x x + /x x 3 + /x x + / x x = Hence, the values of the function grow unbounded on both infinite ends of its domain. 3. intercepts (a) y-intercept Since p(0) = 03 3(0) + =. Hence the graph of p has the y-intercept (0, ). 0 + (b) x-intercept(s) 4. graph We need the roots of the numerator x 3 3x + that are in the domain of p. Since x 3 3x + = (x + )(x x + ) = (x + )(x )(x ) = 0 has the solutions x = and x =, the potential intercepts occur at x = and x =. But only x = is in the domain of p. Hence, the x-intercept is (, 0). We begin by graphing the hole, the intercepts, and the behavior at the infinite ends of the domain. Then we use continuity to complete the graph.
7. (a) If f(x) = 5 and f(x) = 5, what do you know about f(x)? What do you x x + x know about f()? Explain. Solution: Since the it from the right equals the it from the left, we know that f(x) = 5. These its tell us NOTHING about how the function behaves AT x =. x Only how the function behaves NEAR x =. Hence, we know nothing about the value f(), if it even exists. (b) If f(x) = 8, but f(x) does not exist, what do you know about f(x)? x + x x Explain. Solution: We know that the it from the left is not the same as the it from the right. Hence, f(x) 8. x (c) If f(x) =, f(x) = 3, and f(x) =, what do you know about any x x + horizontal and vertical asymptotes of the graph of y = f(x)? Explain. Solution: By the definition of a horizontal asymptote, since f(x) = 3, the graph of y = f(x) has a horizontal asymptote at y = 3. By the definition of a vertical asymptote, since x + f(x) =, the graph of y = f(x) has a vertical asymptote at x =.
8. Sketch the graph of a function with all of the following its and values: f(x) =, f(x) DNE, f(5) = 4, f(0) =, f(x) = 0, f(x) = x 5 x 5 x Solution: There are many possible correct graphs that exhibit all of these behaviors. Only two possibilities are given here.
9. Use the following graph of y = g(x) to answer the questions below: (a) State the domain and range of g in interval notation. Solution: There is a point on the graph corresponding to every x value except x =. Hence, the domain of g is (, ) (, 5]. The range is (, ) (, 3] (b) What is g( )? What is g()? What is g(4)? Solution: Here, looking for the y-values corresponding to the points where x =, x = and x = 4. We see that g( ) =, g() is undefined (as x = not in the domain of g), and g(4) = 3. (c) For what values of x does (i) g(x) =?, (ii) g(x) =?, (iii) g(x) =? Solution: We start by looking as the set of points on the graph of g that intersect the horizontal lines, y =, y = and y =, respectively. For y = we see that y = intersects the graph at the points (, ), (3, ) and the line interval from (, ) to (0, ). Hence, the solutions to the equation g(x) = are the set {x R x 0, x = or x = 3}.
For y =, there is only one point of intersection with the graph; ( 3, ). Hence, g(x) = when x = 3. For y =, there is no point of intersection with the graph. Hence, there are no solutions to the equation g(x) =. (d) Determine all of the following its from the graph of y = g(x): x g(x) x g(x) x + g(x) x 0 g(x) x 4 g(x) x 4 + g(x) x g(x) Solution: g(x) = x g(x) = x x + g(x) = x 0 g(x) does not exist (since the it from the right of x = 0 is and the it from the left of x = 0 is ) x 4 g(x) = g(x) = x 4 + x g(x) = 3