15.44
CURVE SKETCHING (GRAPHS) 0 111
/(*)=^
5/3 -3*
2/3 .
f /(*) = 3*
2 '
3 (U-1).
/'(*) = *
2/3 -2*-"
3 = *-"
3 (*-2) = (*-2)/^.
/"M=i*-"
3 +I*4 " =
2^-4/3^ + j) = i(x + \)r$Tx\ There are critical numbers at 2 and 0. /"(2)>0; hence, there is a relative
minimum at x'= 2, y = 3^4(-|)» -3.2. Near X = 0, f'(x) is positive to the left and negative to the
right, so we have the case { + , -} of the first-derivative test and there is a relative maximum at x = 0, y = 0.
As j _»+«), /(*)-»+=°. As x-*~
x . f( X )->~
x . Note that the graph cuts the x-axis at the solution
x = 5 of 5x - 1 = 0. There is an inflection point at A- = -1, y = -3.6. The graph is concave downward
for x < -1 and concave upward elsewhere. Observe that there is a cusp at the origin. See Fig. 15-33.
15.45
/(x) = 3*
5 -5*
3 + l.
I f'(x) = 15x*-15x
2 = ].5x
2 ( X
2 -l)=l5x:(x-l)(x+l), and /"(*) = 15(4r' - 2x) = 30x(2x2 - 1). The
critical numbers are 0, ±1. /"(1) = 30>0, so there is a relative minimum at x = 1, y = -l. /"(~1) =
-30 < 0, so there is a relative maximum at x = - 1, >' = 3. Near * = 0, f'(x) is negative to the right and
left of x = 0 (since x~ >0 and x
2 -l<0). Thus, we have the case {-,-} of the first-derivative test,
and therefore, there is an inflection point at x = 0, y = l. As JT-»+=C, /(x)->+». As *->-o°,
/(*)-» -oo. There are also inflection points at the solutions of 2x
2 - 1 =0, x = ±V2/2= ±0.7. See Fig.
15-34.
15.46 f( x ) = x'-2 X
2 + l.
I Note that the function is even. f(x) = (x
2 - I)
2 , /'(*) = 4x
3 - 4;e = 4x(Ar
2 - 1) = 4x(x -!)(* + 1), and
f"(x) = 4(3x
2 - I). The critical numbers are 0, ±1. /"(0) = -4<0, so there is a relative maximum at x = 0,
y = l. /"(±1) = 8>0, so there are relative minima at jc = ±l, y = 0. There are inflection points where
3jc2-l=0, x = ±V3/3~Q.6, y= |=0.4 As x^> ±*, f(x)-++«>. See Fig. 15-35.
Fig. 15-35
15.47
f(x) =
I The function is not defined when x
2 <9, that is, for -3<*<3. Observe also that /(-*) = -/(*),
so the graph is symmetric with respect to the origin.
Fig. 15-33
Fig. 15-34