Hence, /"(0) = —2<0, and, therefore, by the second-derivative test, f(x) has a relative maximum at
0. Similarly, f"(2) = 2 > 0, and, therefore, /(*) has a relative minimum at 2.
13.5
Find the critical numbers of f(x) = x
3 - 5x
2 - 8x + 3, and determine whether they yield relative maxima,
relative minima, or inflection points.
f /'(*) = 3*
2 - 10* - 8 = (3x + 2)(x -4). Hence, the critical numbers are x=4 and x=—\. Now,
f"(x) - 6.v - 10. So, /"(4) = 14 > 0, and, by the second-derivative test, there is a relative minimum at x =
4. Similarly, /"(— f) = -14, and, therefore, there is a relative maximum at x = - §.
13.6
Find the critical numbers of /(*) = *(* - I)
3 , and determine whether they yield relative maxima, relative
minima, or inflection points.
f'(x) = x-3(x-l)
2 + (x-lY = (x- l)
2 (3x + x-l) = (x- 1) (4x - 1). So, the critical numbers are x = I
and x=\. Now, f"(x) = (x - I)2 • 4 + 2(x - l)(4x - 1) = 2(x - l)[2(jt - 1) + 4* - 1] = 2(jt - 1)(6* - 3)
= 6(jf - l)(2;c - 1). Thus, /"(J)
= 6(-i)(-j) = ! >0, and, therefore, by the second-derivative test, there is a
relative minimum at *=j. On the other hand /"(I) = 6-0-1 =0, and, therefore, the second-derivative
test is inapplicable. Let us use the first-derivative test. f'(x) - (x - 1)
2
(4* - 1). For jr^l, (A:-I)
2 is
positive. Since 4x — 1 has the value 3 when x = l, 4x —1>0 just to the left and to the right of 1.
Hence,/'(*) is positive both on the left and on the right of x = 1, and this means that we have the case { + , +}.
By the first-derivative test, there is an inflection point at x = 1.
13.7
Find the critical numbers of f(x) = sinx — x, and determine whether they yield relative maxima, relative
minima, or inflection points.
I f'(x) = cos x — 1. The critical numbers are the solutions of cos* = 1, and these are the numbers x =
2irn for any integer n. Now, f"(x) = —sinx. So, f"(2irn) = -sin (Iirn) = -0 = 0, and, therefore, the
second-derivative test is inapplicable. Let us use the first-derivative test. Immediately to the left and right
of x = 2irn, cosjc
13.8
Find the critical numbers of f(x) = (x - I)
2 '
3
and determine whether they yield relative maxima, relative
minima, or inflection points.
I /'(*)= !(x-l)~"
3 = §{l/(x-l)"
3 ]. There are no values of x for which /'(*) = 0, but jt = 1 is a
critical number, since/'(l) is not defined. Try the first-derivative test [which is also applicable when/'(c) is not
defined]. To the left of x = \, (x - 1) is negative, and, therefore,/'(.*) is negative. To the right of x = l,
(jt-1) is positive, and, therefore, f(x) is positive. Thus, the case {-,+} holds, and there is a relative
minimum at x = 1.
13.9
Describe a procedure for finding the absolute maximum and absolute minimum values of a continuous function
f(x) on a closed interval [a, b\.
I Find all the critical numbers of f(x) in [a, b]. List all these critical numbers, c,, c 2 ,.. ., and add the
endpoints a and b to the list. Calculate /(AC) for each x in the list. The largest value thus obtained is the
maximum value of f(x) on [a, b], and the minimal value thus obtained is the minimal value of f(x) on [a, b].
82
CHAPTER 13
13.4
Find the critical numbers of /(*) = x l(x - 1) and determine whether they yield relative maxima, relative
minima, or inflection points.
Hence, the critical numbers are x = 0 and x — 2. [x = 1 is not a critical number because /(I) is not
defined.] Now let us compute f"(x).
0. Similarly, f"(2) = 2 > 0, and, therefore, /(*) has a relative minimum at 2.
13.5
Find the critical numbers of f(x) = x
3 - 5x
2 - 8x + 3, and determine whether they yield relative maxima,
relative minima, or inflection points.
f /'(*) = 3*
2 - 10* - 8 = (3x + 2)(x -4). Hence, the critical numbers are x=4 and x=—\. Now,
f"(x) - 6.v - 10. So, /"(4) = 14 > 0, and, by the second-derivative test, there is a relative minimum at x =
4. Similarly, /"(— f) = -14, and, therefore, there is a relative maximum at x = - §.
13.6
Find the critical numbers of /(*) = *(* - I)
3 , and determine whether they yield relative maxima, relative
minima, or inflection points.
f'(x) = x-3(x-l)
2 + (x-lY = (x- l)
2 (3x + x-l) = (x- 1) (4x - 1). So, the critical numbers are x = I
and x=\. Now, f"(x) = (x - I)2 • 4 + 2(x - l)(4x - 1) = 2(x - l)[2(jt - 1) + 4* - 1] = 2(jt - 1)(6* - 3)
= 6(jf - l)(2;c - 1). Thus, /"(J)
= 6(-i)(-j) = ! >0, and, therefore, by the second-derivative test, there is a
relative minimum at *=j. On the other hand /"(I) = 6-0-1 =0, and, therefore, the second-derivative
test is inapplicable. Let us use the first-derivative test. f'(x) - (x - 1)
2
(4* - 1). For jr^l, (A:-I)
2 is
positive. Since 4x — 1 has the value 3 when x = l, 4x —1>0 just to the left and to the right of 1.
Hence,/'(*) is positive both on the left and on the right of x = 1, and this means that we have the case { + , +}.
By the first-derivative test, there is an inflection point at x = 1.
13.7
Find the critical numbers of f(x) = sinx — x, and determine whether they yield relative maxima, relative
minima, or inflection points.
I f'(x) = cos x — 1. The critical numbers are the solutions of cos* = 1, and these are the numbers x =
2irn for any integer n. Now, f"(x) = —sinx. So, f"(2irn) = -sin (Iirn) = -0 = 0, and, therefore, the
second-derivative test is inapplicable. Let us use the first-derivative test. Immediately to the left and right
of x = 2irn, cosjc
13.8
Find the critical numbers of f(x) = (x - I)
2 '
3
and determine whether they yield relative maxima, relative
minima, or inflection points.
I /'(*)= !(x-l)~"
3 = §{l/(x-l)"
3 ]. There are no values of x for which /'(*) = 0, but jt = 1 is a
critical number, since/'(l) is not defined. Try the first-derivative test [which is also applicable when/'(c) is not
defined]. To the left of x = \, (x - 1) is negative, and, therefore,/'(.*) is negative. To the right of x = l,
(jt-1) is positive, and, therefore, f(x) is positive. Thus, the case {-,+} holds, and there is a relative
minimum at x = 1.
13.9
Describe a procedure for finding the absolute maximum and absolute minimum values of a continuous function
f(x) on a closed interval [a, b\.
I Find all the critical numbers of f(x) in [a, b]. List all these critical numbers, c,, c 2 ,.. ., and add the
endpoints a and b to the list. Calculate /(AC) for each x in the list. The largest value thus obtained is the
maximum value of f(x) on [a, b], and the minimal value thus obtained is the minimal value of f(x) on [a, b].
82
CHAPTER 13
13.4
Find the critical numbers of /(*) = x l(x - 1) and determine whether they yield relative maxima, relative
minima, or inflection points.
Hence, the critical numbers are x = 0 and x — 2. [x = 1 is not a critical number because /(I) is not
defined.] Now let us compute f"(x).