Sketching Reciprocals of Polynomials
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1 Graphs of Reciprocals of Polynomials Sketching Reciprocals of Polynomials The reciprocal of a polynomial is of the form f(x) =, where p(x) is a polynomial. The roots of the denominator p(x) p(x) determine the vertical asymptotes of the graph of f, which itself has no roots. The x-axis is the horizontal asymptote of the graph of f. From the Power Rule Doug MacLean f (x) = ( )(p(x)) p (x) = p (x) (p(x)), and from the Quotient Rule, f (x) = (p(x)) p (x) p(x)p (x)p (x) (p(x)) 4 = (p (x)) p(x)p (x) (p(x)), so the interesting points occur at the roots of the three polynomials p(x), p (x), and (p (x)) p(x)p (x). In our examples we will often first graph a specific function, and then look at the general class of functions of which it is a special case. Note: For most problems, the graph can be viewed interactively using Java applets with Netscape Communicator or Internet Explorer. Colour Conventions: The graph of the function being studied is drawn in black, that of the first derivative is in red, and that of the second derivative is in blue. The denominator s graph is drawn in purple. The asymptotes are drawn in green, except when they coincide with the axes, which are black.
2 Graphs of Reciprocals of Polynomials Example : f(x) = x a Doug MacLean Step : f (x) = (x a) < and f (x) = (x a). The only interesting value for f is a. Step : Put these values of x into increasing order. a. Step : Put together as good a table as you can showing the signs of f (x) and f (x) on the intervals into which the interesting values divide the domain of f. x (, a) a (a, ) f (x) UND + f (x) UND f(x) UND Step 4: half-frowns. Plot the interesting points and connect them with curves which are either left or right half-smiles or But now we have a problem: the only interesting x-value, a, doesn t have a corresponding y-value because is not in the domain of f! Even though the function is not defined at, we can compute the left- and right- hand limits there: lim f(x) = lim x a x a x =, and lim f(x) = lim x a x a + x =+.
3 Graphs of Reciprocals of Polynomials We can also compute the limits at and + : lim f(x) = lim x x x =, and lim f(x) = lim x + x + x =. We can also easily see that f(a ) = and f(a) =. We can represent these facts in a modified table, using a bit of shorthand notation: x (,a) a a + (a, ) f (x) a + + a f (x) f(x) f(a ) = f(a+ ) = We plot the graph with a = : y x
4 4 Graphs of Reciprocals of Polynomials Example : f(x) = x = (x )(x + ) Doug MacLean Step : f x (x) = (x ) and f (x) = 6x + (x ). The interesting values for f are,, and. Step : Put these values of x into increasing order.,,. Step : Put together as good a table as you can showing the signs of f (x) and f (x) on the intervals into which the interesting values divide the domain of f. x (, ) (, ) (, ) (, ) f (x) + UND UND + f (x) + UND + UND f(x) + UND UND + Step 4: half-frowns. Plot the interesting points and connect them with curves which are either left or right half-smiles or Even though the function is not defined at ±, we can compute the left- and right- hand limits there: lim x f(x) = lim f(x) = lim x x x lim =, and x x lim x f(x) = lim x + x =. =, and lim f(x) = lim x x + x =.
5 Graphs of Reciprocals of Polynomials 5 We can also compute the limits at and + : lim f(x) = lim x x We use a modified table: =, and lim f(x) = lim x x + x + x =. x (, ) + (, ) (, ) + (, ) f (x) f (x) f(x) y x
6 6 Graphs of Reciprocals of Polynomials Example : f(x) = x (a + b)x ab =, where a b. We assume a<b. (x a)(x b) Doug MacLean Step : f x (a + b) (x) = (x a) (x b) and f (x) = x (a + b)x + a + ab + b. since the discriminant of the numerator (x a) (x b) is (a b) < The interesting values for f are a, b and a + b. Step : Put these values of x into increasing order. a, a + b,b. Step : Put together as good a table as you can showing the signs of f (x) and f (x) on the intervals into which the interesting values divide the domain of f. x (,a) a (, a + b ) a + b ( ) a + b,b b (b, ) f (x) + UND UND + f (x) + UND + UND f(x) + UND Step 4: half-frowns. 4 (b a) UND + Plot the interesting points and connect them with curves which are either left or right half-smiles or Even though the function is not defined at a and b, we can compute the left- and right- hand limits there:
7 Graphs of Reciprocals of Polynomials 7 lim f(x) = lim x a x a (x a)(x b) lim f(x) = lim x b x b (x a)(x b) We can also compute the limits at and + : lim f(x) = lim x x We use a modified table: =, and lim f(x) = lim x a x a + (x a)(x b) =. =, and lim f(x) = lim x b x b + (x a)(x b) =. =, and lim f(x) = lim (x a)(x b) x + x + ( x (,a) a a + a, a + b ) (x a)(x b) =. a + b ( ) a + b,b b b + (b, ) f (x) f (x) f(x) (b a) + +
8 8 Graphs of Reciprocals of Polynomials Graph for Example : y Doug MacLean x
9 Graphs of Reciprocals of Polynomials 9 Sketching y = f(x) = p(x) = (x r ) (x r ) (x r n ) This graph will have vertical asymptotes x = r, x = r,,x = r n. We assume that r <r < <r n. Since f (x) = p (x) (p(x)), f (x) = requires p (x) =, so the First Derivative Test tells us that the relative maxima of f correspond to the relative minima of p, and vice versa. Example 4: f(x) = x (a + b + c)x + (ab + ac + bc)x abc =, where a<b<c. (x a)(x b)(x c) Doug MacLean Step : p(x) = x (a + b + c)x + (ab + ac + bc)x abc, p (x) = x (a + b + c)x + (ab + ac + bc), p (x) = 6x (a + b + c). f (x) = x (a + b + c)x + (ab + ac + bc) (x a) (x b) (x c) and f (x) = (p (x)) p (x)p(x) (x a) (x b) (x c). p (x) = ifx = (a + b + c) ± 4(a + b + c) 4 (ab + ac + bc) 6 a + b + c ± (a + b + c) (ab + ac + bc) = a + b + c ± a + b + c ab ac bc = (a + b + c) ± (a + b + c) (ab + ac + bc) 6 =
10 Graphs of Reciprocals of Polynomials So, some of the interesting values for f are thus a, b, c, and a + b + c ± a + b + c ab ac bc. It is clear that the calculation of the roots of the second derivative would be very difficult. Step : Put these values of x into increasing order. By Rolle s Theorem, we know that the roots of p (x) lie between those of p(x), so we have a< a + b + c a + b + c ab ac bc <b< a + b + c + a + b + c ab ac bc <c. As a matter of fact, we can conclude that a + b + c ab + ac + bc because Rolle s Theorem tells us that p (x) must have exactly two roots. For brevity, we denote these two roots by α = a + b + c a + b + c ab ac bc and β = a + b + c + a + b + c ab ac bc. Step : Put together as good a table as you can showing the signs of f (x) and f (x) on the intervals into which the interesting values divide the domain of f. x (,a) a a + (a, α) α (α, b) b b + (b, β) β (β, c) c c + (c, ) f (x) f (x) f(x) Step 4: half-frowns. Plot the interesting points and connect them with curves which are either left or right half-smiles or
11 Graphs of Reciprocals of Polynomials In the following diagram, we have used the values a =, b =, and c = : y x
12 Graphs of Reciprocals of Polynomials Example 5 y = f(x) = x (x + ) Solution: f x + (x) = x (x + ), f (x) = 6 x + 8x + 9, so the important x-values are,, and. x 4 (x + ) We construct a table showing the signs of the first and second derivatives: Doug MacLean (, ) + (, ) (, ) + (, ) f (x) f (x) + + f(x) y x - -4
13 Graphs of Reciprocals of Polynomials, Example 6 y = f(x) = Solution: f (x) = ( (x a) (x b) (x a) (x b) ) a)(x b) + (x a) ((x a) = (x = (x b)) (x a) 4 (x b) (x b) + x a x (a + b) = (x a) (x b) (x a) (x b) Doug MacLean f (x) = (x a) (x b) (x (a + b)) ( (x a) (x b) ) (x (a + b)) ((x a) (x b) ) = (x a) (x b) () [ (x a) (x b) + (x a) (x b) ] (x (a + b)) = (x a) 6 (x b) 4 (x a)(x b) [(x b) + (x a)] (x (a + b)) = (x a) 4 (x b) (x a)(x b) [5x (a + b)] (x (a + b)) = (x a) 4 (x b) x (a + b)x + ab [ 5x (a + 5b)x + (a + b)(a + b) ] = (x a) 4 (x b) x + bx + a + 4ab + 5b (x a) 4 (x b) Since the discriminant of the numerator, 4b 4()(a + 4ab + 5b ) = 4( 4a 48ab 59b ) = 4( 4a 48ab 4b 5b ) = 4( 4(a + b) 5b )<, the numerator has no roots, and thus f has no inflection points. so the important x-values are a, a + b, and b. We construct a table showing the signs of the first and second derivatives: ( (,a) a a + a, a + b ) ( ) a + b a + b,b b b + (b, ) f (x) + + f (x) + + f(x) + +
14 4 Graphs of Reciprocals of Polynomials y x
15 Graphs of Reciprocals of Polynomials 5 Example 7 y = f(x) = ax + bx + c, b 4ac < Solution: The domain of f is (, ), since the denominator has no roots. f (x) = x = b b ±, and is always defined if x = a b 4ac. a ax + b (ax + bx + c) = if Doug MacLean f (x) = (ax + b) a(ax + bx + c) = 8a x + 8abx + b a x abx ac = a x + abx + b ac (ax + bx + c) (ax + bx + c) (ax + bx + c) whose numerator has discriminant (ab) 4(a )(b ac) = 9a b a b +a c = a c a b = a (4ac b )> so the important x-values are b (4ac b, b + (4ac b, and b. a a a We let D = (4ac b. We construct a table showing the signs of the first and second derivatives: (, b D ) ( b D b D a a a, b ) ( b b a a a, b + D ) b + D ( ) b+d a a a f (x) + + f (x) f(x)
16 6 Graphs of Reciprocals of Polynomials We sketch the graph for a =, b =, and c =. y 4 Doug MacLean x - -
17 Graphs of Reciprocals of Polynomials 7 Example 8 y = f(x) = ax + bx + c, b 4ac > Solution: Since the roots of ax + bx + c are b ± b 4ac, the domain of f is ( a b b 4ac, b + ) ( ) b 4ac b + b 4ac,. a a a (, b ) b 4ac a Doug MacLean f ax + b b b ± b 4ac (x) = = if x =, and is undefined if x =. (ax + bx + c) a a f (x) = (ax + b) a(ax + bx + c) = 8a x + 8abx + b a x abx ac = a x + abx + b ac (ax + bx + c) (ax + bx + c) (ax + bx + c) whose numerator has discriminant (ab) 4(a )(b ac) = 9a b a b +a c = a c a b = a (4ac b )< so the important x-values are again b (4ac b, b + (4ac b, and b. a a a We let D = (4ac b. We construct a table showing the signs of the first and second derivatives: (, b D ) ( b D b D a a a, b ) ( b b a a a, b + D ) b + D ( ) b+d a a a f (x) + UND UND + f (x) + UND + UND f(x) + UND UND +
18 8 Graphs of Reciprocals of Polynomials We sketch the graph for a =, b =, and c =. Doug MacLean y x
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