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) = x
2 - 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 + I5x
2 + 6x + 1.
I f'(x) = 3x
2 +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) = x
4 + I8x
3 + UOx
2 + x + l.
I f'(x) = 4x
j + 54x
2 +240x + l,
f"(x) = \2x
2 + W8x + 240= 12(x
2 + 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, therefore, so is /"(•*). For -5 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/(2x - 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) point, since f(x) is not defined when x = |.
15.5
f(x) = 5x
4 - x".
I f'(x) = 2Qx*-5x\ and /"(*) = 6Qx
2 -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-Jc<0, /"(AT) 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 + x
2 .
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) = A4 - ISA:
2 + 9.
I /'W = 4Ar
3 -36A: = 4A:(A;
2 -9) = 4A:(A:-3)(A: + 3). f(x) = 12;t
2 - 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.
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