CHAPTER 15
Curve Sketching (Graphs)
When sketching a graph, show all relative extrema, inflection points, and asymptotes; indicate concavity; and
suggest the behavior at infinity.
In Problems 15.1 to 15.5, determine the intervals where the graphs of the following functions are concave
upward and where they are concave downward. Find all inflection points.
15.1 f(x) = x2 - x + 12.
I f ' ( x ) = 2x — l, f"(x) = 2. Since the second derivative is always positive, the graph is always concave
upward and there are no inflection points.
15.2 /(*) = A:3 + I5x2 + 6x + 1.
I f ' ( x ) = 3x2+3Qx + 6, f"(x) = 6x + 30 = 6(x + 5). Thus, f"(x)>0 when jc>-5; hence, the graph is
concave upward for ;t>-5. Since f"(x)<0 for x<-5, the graph is concave downward for x<-5.
Hence, there is an inflection point, where the concavity changes, at (5,531).
15.3 /(AT) = x4 + I8x3 + UOx2 + x + l.
I f ' ( x ) = 4xj+ 54x2+240x + l, f"(x) = \2x2 + W8x + 240= 12(x2 + 9x + 20) = U(x + 4)(x + 5). Thus,
the important points are x = -4 and x = -5. For x > -4, x + 4 and x + 5 are positive, and, there-
fore, so is /"(•*). For -5<A:<-4, x + 4 is negative and x + 5 is positive; he nee, f"(x) is negative. For
x<-5, both x+4 and x + 5 are negative, and, therefore,/"(A:) is positive. Therefore, the graph is
concave upward for x > -4 and for A: < -5. The graph is concave downward for -5 < x < -4. Thus,
the inflection points are (-4,1021) and (-5,1371).
15.4 f(x) = x/(2x-l).
I /(*) = [ ( * - i ) + | ] / [ 2 ( * - i ) ] = i { l + [l/(2*-l)]}. Hence, /'(*) = i[-l/(2* - I) 2 ] - 2 = -l/(2x - I)2.
Then /"(*) = [ l / ( 2 x - I) 3 ] - 2 = 2/(2x - I)3. For x>$, 2* - 1 >0, f"(x)>0, and the graph is concave
upward. For x<\, 2x — 1<0, /"(AC)<O, and the graph is concave downward. There is no inflection
point, since f(x) is not defined when x = |.
15.5 f(x) = 5x4 - x".
I f ' ( x ) = 2Qx*-5x\ and /"(*) = 6Qx2 -20x* = 20x\3 - x). So, for 0 < * < 3 and for *<0, 3-
;c>0, /"(A-)>O, and the graph is concave upward. For jc>3, 3 - J c < 0 , /"(AT)<O, and the graph is
concave downward. There is an inflection point at (3,162). There is no inflection point at x = 0; the graph
is concave upward for x < 3.
For Problems 15.6 to 15.10, find the critical numbers and determine whether they yield relative maxima, relative
minima, inflection points, or none of these.
15.6 f(x) = 8 - 3x + x2.
I f ' ( x ) = -3 + 2x, f"(x) = 2. Setting -3 + 2A: = 0, we find that x = \ is a critical number. Since
/"(AC) = 2>0, the second-derivative test tells us that there is a relative minimum at x= \.
15.7 f(x) = A-4 - ISA:2 + 9.
I /'W = 4Ar 3 -36A: = 4A:(A; 2 -9) = 4A:(A:-3)(A: + 3). f(x) = 12;t2 - 36 = 12(A:2 - 3). The critical numbers
are 0, 3, -3. /"(O) =-36<0; hence, x = 0 yields a relative maximum. /"(3) = 72>0; hence, x = 3
yields a relative minimum. /"(-3) = 72>0; hence x = -3 yields a relative minimum. There are inflection
points at x = ±V5, y = -36.
100
, CURVE SKETCHING (GRAPHS) D 101
15.8 f(x) = x-5x -8*+ 3.
I /'(*) = 3* 2 -10jt-8 = (3;c + 2)(.x-4). /"(*) = 6x - 10. The critical numbers are x = -\ and A: = 4.
/"(-§)=-14 <0; hence, * = - § yields a relative maximum. /"(4)=14>0; hence, x = 4 yields a
relative minimum. There is an inflection point at x = f .
15.9 f(x) = x2/(x-l).
So, / ' W = l - l / ( x - l ) 2 , /"(x) = 2/(^-l) 3 .
Thus, the critical numbers are the solutions of l = l / ( j e — 1 ) , (*-l) = 1, * — 1 = ±1, x=Q or x = 2.
/"(0) = -2<0; thus, jc=0 yields a relative maximum. /"(2) = 2>0; thus, A: = 2 yields a relative
minimum.
15.10 f(x) = x2/(x2 + l).
I f(x) = 1- l/(jc 2 + 1). So, f ' ( x ) = 2x/(x2 + I)2. The only critical number is x = 0. Use the first-deriva-
tive test. To the right of 0, f'(x)>0, and to the left of 0, /'(*)<0. Thus, we have the case {-,+}, and,
therefore, x = 0 yields a relative minimum. Using the quotient rule, f"(x) = 2(1 - 3jt 2 )/(;t 2 + I)3. So,
there are inflection points at x = ± 1 /V3, y = 1.
In Problems 15.11 to 15.19, sketch the graph of the given function.
15.11 f(x) = (x2 - I)3.
2
| f(x) = 3(x - i)2 . 2x = 6x(x2 - I)2 = 6x(x - l)\x + I)2. There are three critical numbers 0,1, and -1.
At x=Q, f'(x)>0 to the right of 0, and /'(*)< 0 to the left of 0. Hence, we have the case {-,+} of
the first derivative test; thus x = 0 yields a relative minimum at (0, -1). For both x = l and x = —I,
f ' ( x ) has the same sign to the right and to the left of the critical number; therefore, there are inflection points at
(1,0) and (-1,0). When x-»±=°, /(*)-»+».
It is obvious from what we have of the graph in Fig. 15-1 so far that there must be inflection points between
x=— 1 and x = 0, and between x = Q and jc = l. To find them, we compute the second derivative:
/"(*) = 6(x - l)2(x + I) 2 + I2x(x - l)2(x + 1) + 12*(x - l)(x + I) 2
= 6(x — l)(x + l)[(x — l)(x + 1) + 2x(x - 1) + 2x(x + 1)]
=6(x-1)(x+1)(5x2-1)
Hence, the inflection points occur when 5x2 - 1 = 0 . x2 = \,
—0.51. The graph is in Fig. 15-1.
Fig. 15-1 (2, -5) Fig. 15-2
15.12 f(x) = x3 - 2x2 - 4x + 3.
I /'(*) = 3* 2 -4A:-4=(3.x + 2)(;c-2). /"(*) = 6x - 4 = 6(x - 1). The critical numbers are * = -f
and x = 2. /"(-§)=-8<0; hence, there is a relative maximum at Jt = -|, > > = ^ = 4 . 5 . /"(2) = 8>
0; so there is a relative minimum at A: = 2, y = — 5. As jc-»+x, /(*)—* +°°- As x—»—»,
/(j:)-» -oo. To find the inflection point(s), we set f"(x) = 6x - 4 = 0, obtaining * = § , y--yi'a -0.26.
The graph is shown in Fig. 15-2.
, 102 0 CHAPTER 15
15.13 f(x) = x(x - 2)2.
| f'(x) = x.2(x-2) + (x-2)2 = (x-2)(3x-2), and f"(x) = (x -2)-3 + (3* -2) = 6* -8. The critical
numbers are jc = 2 and j c = | . /"(2) = 4>0; hence, there is a relative minimum at Jt = 2, y = 0 .
/"(|) = -4 < Q; hence, there is a relative maximum at * = |, y = H ~ 1.2. There is an inflection point
where /"(*) = 6* - 8 = 0, x = § , y = $ ~ 0 . 6 . As x -»+«>, /(*)-»+00. As x-»-oo, /(*)-»-°°.
Notice that the graph intersects the x-axis at x = 0 and x-2. The graph is shown in Fig. 15-3.
Fig. 15-3 Fig. 15-4
15.14 f(x) = x*+4x3.
I f'(x) = 4x3 + 12x2 = 4x2(x + 3) and /"(*) = 12x2 + 24x = 12X* + 2). The critical numbers are x = 0
and x = -3. /"(O) = 0, so we have to use the first-derivative test: f ' ( x ) is positive to the right and left of 0;
hence, there is an inflection point at x = Q, y = 0. /"(-3) = 36>0; hence, there is a relative minimum at
x=—3, y = —27. Solving f"(x) = 0, we see that there is another inflection point at x = — 2, y = —16.
Since f(x) = x}(x + 4), the graph intersects the AT-axis only at x = 0 and x =
—4. As x—* ±o°, /(*)—»+<». The graph is shown in Fig. 15-4.
15.15 /(A:) = 3A:5 - 2(k3.
I /'(A:) = 15A:4-60jc2 = 15xV-4) = 15A:2(^-2)(A: + 2), and /"(*) = 60x3 - 120x = 6CU(;r - 2)i =
60A:(A: - V2)(x + V2). The critical numbers are 0,2, -2. /"(0) = 0. So, we must use the first-derivative
test for x = 0. f ' ( x ) is negative to the right and left of x = 0; hence, we have the {-,-} case, and there is
an inflection point at x = 0, y = 0. For x = 2, f"(2) = 240 > 0; thus, there is a relative minimum at
x = 2, y = -64. Similarly, f"(-2) =-240<0, so there is a relative maximum at x = -2, y = (A. There
are also inflection points at x = V2, y = -28V5« -39.2, and at x =-V2, >• = 28V2 = 39.2. As
A:-*+=C, /(X)-*+M. As JT-»-<», /(*)-»-<». See Fig. 15-5.
Fig. 15-5
15.16
' /'W = s(jf - l)" 2/ "\ and /"(*) = ~ i ( x ~ V'*- The only critical number is x = l, where/'(*) is not
denned. /(1) = 0. f ' ( x ) is positive to the right and left of x = \. f"(x) is negative to the right of j c = l ; so
the graph is concave downward for x > 1. f"(x) is positive to the left of x = 1; hence, the graph is concave
upward for x<l and there is an inflection point at x = \. As *—»+°°, f(x)—» +«, and, as JT—»—»,
/W^-=o. See Fig. 15-6.
Curve Sketching (Graphs)
When sketching a graph, show all relative extrema, inflection points, and asymptotes; indicate concavity; and
suggest the behavior at infinity.
In Problems 15.1 to 15.5, determine the intervals where the graphs of the following functions are concave
upward and where they are concave downward. Find all inflection points.
15.1 f(x) = x2 - x + 12.
I f ' ( x ) = 2x — l, f"(x) = 2. Since the second derivative is always positive, the graph is always concave
upward and there are no inflection points.
15.2 /(*) = A:3 + I5x2 + 6x + 1.
I f ' ( x ) = 3x2+3Qx + 6, f"(x) = 6x + 30 = 6(x + 5). Thus, f"(x)>0 when jc>-5; hence, the graph is
concave upward for ;t>-5. Since f"(x)<0 for x<-5, the graph is concave downward for x<-5.
Hence, there is an inflection point, where the concavity changes, at (5,531).
15.3 /(AT) = x4 + I8x3 + UOx2 + x + l.
I f ' ( x ) = 4xj+ 54x2+240x + l, f"(x) = \2x2 + W8x + 240= 12(x2 + 9x + 20) = U(x + 4)(x + 5). Thus,
the important points are x = -4 and x = -5. For x > -4, x + 4 and x + 5 are positive, and, there-
fore, so is /"(•*). For -5<A:<-4, x + 4 is negative and x + 5 is positive; he nee, f"(x) is negative. For
x<-5, both x+4 and x + 5 are negative, and, therefore,/"(A:) is positive. Therefore, the graph is
concave upward for x > -4 and for A: < -5. The graph is concave downward for -5 < x < -4. Thus,
the inflection points are (-4,1021) and (-5,1371).
15.4 f(x) = x/(2x-l).
I /(*) = [ ( * - i ) + | ] / [ 2 ( * - i ) ] = i { l + [l/(2*-l)]}. Hence, /'(*) = i[-l/(2* - I) 2 ] - 2 = -l/(2x - I)2.
Then /"(*) = [ l / ( 2 x - I) 3 ] - 2 = 2/(2x - I)3. For x>$, 2* - 1 >0, f"(x)>0, and the graph is concave
upward. For x<\, 2x — 1<0, /"(AC)<O, and the graph is concave downward. There is no inflection
point, since f(x) is not defined when x = |.
15.5 f(x) = 5x4 - x".
I f ' ( x ) = 2Qx*-5x\ and /"(*) = 6Qx2 -20x* = 20x\3 - x). So, for 0 < * < 3 and for *<0, 3-
;c>0, /"(A-)>O, and the graph is concave upward. For jc>3, 3 - J c < 0 , /"(AT)<O, and the graph is
concave downward. There is an inflection point at (3,162). There is no inflection point at x = 0; the graph
is concave upward for x < 3.
For Problems 15.6 to 15.10, find the critical numbers and determine whether they yield relative maxima, relative
minima, inflection points, or none of these.
15.6 f(x) = 8 - 3x + x2.
I f ' ( x ) = -3 + 2x, f"(x) = 2. Setting -3 + 2A: = 0, we find that x = \ is a critical number. Since
/"(AC) = 2>0, the second-derivative test tells us that there is a relative minimum at x= \.
15.7 f(x) = A-4 - ISA:2 + 9.
I /'W = 4Ar 3 -36A: = 4A:(A; 2 -9) = 4A:(A:-3)(A: + 3). f(x) = 12;t2 - 36 = 12(A:2 - 3). The critical numbers
are 0, 3, -3. /"(O) =-36<0; hence, x = 0 yields a relative maximum. /"(3) = 72>0; hence, x = 3
yields a relative minimum. /"(-3) = 72>0; hence x = -3 yields a relative minimum. There are inflection
points at x = ±V5, y = -36.
100
, CURVE SKETCHING (GRAPHS) D 101
15.8 f(x) = x-5x -8*+ 3.
I /'(*) = 3* 2 -10jt-8 = (3;c + 2)(.x-4). /"(*) = 6x - 10. The critical numbers are x = -\ and A: = 4.
/"(-§)=-14 <0; hence, * = - § yields a relative maximum. /"(4)=14>0; hence, x = 4 yields a
relative minimum. There is an inflection point at x = f .
15.9 f(x) = x2/(x-l).
So, / ' W = l - l / ( x - l ) 2 , /"(x) = 2/(^-l) 3 .
Thus, the critical numbers are the solutions of l = l / ( j e — 1 ) , (*-l) = 1, * — 1 = ±1, x=Q or x = 2.
/"(0) = -2<0; thus, jc=0 yields a relative maximum. /"(2) = 2>0; thus, A: = 2 yields a relative
minimum.
15.10 f(x) = x2/(x2 + l).
I f(x) = 1- l/(jc 2 + 1). So, f ' ( x ) = 2x/(x2 + I)2. The only critical number is x = 0. Use the first-deriva-
tive test. To the right of 0, f'(x)>0, and to the left of 0, /'(*)<0. Thus, we have the case {-,+}, and,
therefore, x = 0 yields a relative minimum. Using the quotient rule, f"(x) = 2(1 - 3jt 2 )/(;t 2 + I)3. So,
there are inflection points at x = ± 1 /V3, y = 1.
In Problems 15.11 to 15.19, sketch the graph of the given function.
15.11 f(x) = (x2 - I)3.
2
| f(x) = 3(x - i)2 . 2x = 6x(x2 - I)2 = 6x(x - l)\x + I)2. There are three critical numbers 0,1, and -1.
At x=Q, f'(x)>0 to the right of 0, and /'(*)< 0 to the left of 0. Hence, we have the case {-,+} of
the first derivative test; thus x = 0 yields a relative minimum at (0, -1). For both x = l and x = —I,
f ' ( x ) has the same sign to the right and to the left of the critical number; therefore, there are inflection points at
(1,0) and (-1,0). When x-»±=°, /(*)-»+».
It is obvious from what we have of the graph in Fig. 15-1 so far that there must be inflection points between
x=— 1 and x = 0, and between x = Q and jc = l. To find them, we compute the second derivative:
/"(*) = 6(x - l)2(x + I) 2 + I2x(x - l)2(x + 1) + 12*(x - l)(x + I) 2
= 6(x — l)(x + l)[(x — l)(x + 1) + 2x(x - 1) + 2x(x + 1)]
=6(x-1)(x+1)(5x2-1)
Hence, the inflection points occur when 5x2 - 1 = 0 . x2 = \,
—0.51. The graph is in Fig. 15-1.
Fig. 15-1 (2, -5) Fig. 15-2
15.12 f(x) = x3 - 2x2 - 4x + 3.
I /'(*) = 3* 2 -4A:-4=(3.x + 2)(;c-2). /"(*) = 6x - 4 = 6(x - 1). The critical numbers are * = -f
and x = 2. /"(-§)=-8<0; hence, there is a relative maximum at Jt = -|, > > = ^ = 4 . 5 . /"(2) = 8>
0; so there is a relative minimum at A: = 2, y = — 5. As jc-»+x, /(*)—* +°°- As x—»—»,
/(j:)-» -oo. To find the inflection point(s), we set f"(x) = 6x - 4 = 0, obtaining * = § , y--yi'a -0.26.
The graph is shown in Fig. 15-2.
, 102 0 CHAPTER 15
15.13 f(x) = x(x - 2)2.
| f'(x) = x.2(x-2) + (x-2)2 = (x-2)(3x-2), and f"(x) = (x -2)-3 + (3* -2) = 6* -8. The critical
numbers are jc = 2 and j c = | . /"(2) = 4>0; hence, there is a relative minimum at Jt = 2, y = 0 .
/"(|) = -4 < Q; hence, there is a relative maximum at * = |, y = H ~ 1.2. There is an inflection point
where /"(*) = 6* - 8 = 0, x = § , y = $ ~ 0 . 6 . As x -»+«>, /(*)-»+00. As x-»-oo, /(*)-»-°°.
Notice that the graph intersects the x-axis at x = 0 and x-2. The graph is shown in Fig. 15-3.
Fig. 15-3 Fig. 15-4
15.14 f(x) = x*+4x3.
I f'(x) = 4x3 + 12x2 = 4x2(x + 3) and /"(*) = 12x2 + 24x = 12X* + 2). The critical numbers are x = 0
and x = -3. /"(O) = 0, so we have to use the first-derivative test: f ' ( x ) is positive to the right and left of 0;
hence, there is an inflection point at x = Q, y = 0. /"(-3) = 36>0; hence, there is a relative minimum at
x=—3, y = —27. Solving f"(x) = 0, we see that there is another inflection point at x = — 2, y = —16.
Since f(x) = x}(x + 4), the graph intersects the AT-axis only at x = 0 and x =
—4. As x—* ±o°, /(*)—»+<». The graph is shown in Fig. 15-4.
15.15 /(A:) = 3A:5 - 2(k3.
I /'(A:) = 15A:4-60jc2 = 15xV-4) = 15A:2(^-2)(A: + 2), and /"(*) = 60x3 - 120x = 6CU(;r - 2)i =
60A:(A: - V2)(x + V2). The critical numbers are 0,2, -2. /"(0) = 0. So, we must use the first-derivative
test for x = 0. f ' ( x ) is negative to the right and left of x = 0; hence, we have the {-,-} case, and there is
an inflection point at x = 0, y = 0. For x = 2, f"(2) = 240 > 0; thus, there is a relative minimum at
x = 2, y = -64. Similarly, f"(-2) =-240<0, so there is a relative maximum at x = -2, y = (A. There
are also inflection points at x = V2, y = -28V5« -39.2, and at x =-V2, >• = 28V2 = 39.2. As
A:-*+=C, /(X)-*+M. As JT-»-<», /(*)-»-<». See Fig. 15-5.
Fig. 15-5
15.16
' /'W = s(jf - l)" 2/ "\ and /"(*) = ~ i ( x ~ V'*- The only critical number is x = l, where/'(*) is not
denned. /(1) = 0. f ' ( x ) is positive to the right and left of x = \. f"(x) is negative to the right of j c = l ; so
the graph is concave downward for x > 1. f"(x) is positive to the left of x = 1; hence, the graph is concave
upward for x<l and there is an inflection point at x = \. As *—»+°°, f(x)—» +«, and, as JT—»—»,
/W^-=o. See Fig. 15-6.
