Solutions, Section 4.5
31
√
x 3 − 3x2 . f (x) = 0 for x = −1, x = 0 and x = 1, however, x = 0 , f (x) = √ +3 2 x(x2 + 3)2 and x = −1 are outside the interval, thus f has a maximum of 1/4 at x = 1 (ďŹ rst derivative test). There is no minimum.
4.5.11 f (x) =
x2
1 − x2 x (x) = . f (x) = 0 when x = −1 and x = 1, however, x = −1 is ; f x2 + 1 (x2 + 1)2 outside the interval. f (0) = 0; f (1) = 1/2; and f (2) = 2/5, so f has a maximum of 1/2 at x = 1 and a minimum of 0 at x = 0.
4.5.12 (x) =
1 x so there are no critical points. f (0) = 0, thus, for x in [0, +∞); f (x) = 1+x (1 + x)2 f has a minimum of 0 at x = 0. There is no maximum.
4.5.13 f (x) =
2x − 1 1 . f (x) = 0 when x = 1/2; f (x) does not exist at x = 0 , f (x) = x − x2 (x − x2 )2 or x = 1, however, both of these values are outside the interval. f has a minimum of +4 at x = 1/2, there is no maximum.
4.5.14 f (x) =
4.5.15 f (x) =
x2 ,
x<0
3
xâ&#x2030;Ľ0
x ,
f (x) =
2x,
x<0
2
x>0
3x ,
f (x) = 0 when x = 0 which corresponds to a minimum value (ďŹ rst derivative test), there is no maximum.   x < â&#x2C6;&#x2019;1 x < â&#x2C6;&#x2019;1    x â&#x2C6;&#x2019; 1,  â&#x2C6;&#x2019;1, 2 1 â&#x2C6;&#x2019; x , â&#x2C6;&#x2019;1 â&#x2030;¤ x â&#x2030;¤ 1 â&#x2C6;&#x2019;2x, â&#x2C6;&#x2019;1 < x < 1 4.5.16 f (x) = f (x) =     x â&#x2C6;&#x2019; 1, x>1 1, x>1 f (x) = 0 when x = 0, f (x) does not exist at x = â&#x2C6;&#x2019;1 or x = 1. f (â&#x2C6;&#x2019;2) = 1; f (â&#x2C6;&#x2019;1) = 0; f (0) = 1; f (1) = 0; f (2) = 1, thus, f has a maximum of 1 at x = â&#x2C6;&#x2019;2, x = 0, and x = 2; f has a minimum of 0 at x = â&#x2C6;&#x2019;1 and x = 1. 1 â&#x2C6;&#x2019; x2 , x < 0 â&#x2C6;&#x2019;2x, x < 0 4.5.17 f (x) = f (x) = 3 3x2 , x > 0 x â&#x2C6;&#x2019; 1, x â&#x2030;Ľ 0 f (x) = 0 when x = 0. f (â&#x2C6;&#x2019;2) = â&#x2C6;&#x2019;3, f (0) = â&#x2C6;&#x2019;1, f (1) = 0, thus, f has a maximum of 0 at x = 1 and a minimum of â&#x2C6;&#x2019;3 at x = â&#x2C6;&#x2019;2. â&#x2C6;&#x2019;2, x < 3/2 3 â&#x2C6;&#x2019; 2x, x â&#x2030;¤ 3/2 4.5.18 f (x) = 3 â&#x2C6;&#x2019; 2x = f (x) = 2, x > 3/2 â&#x2C6;&#x2019;3 + 2x, x > 3/2 3 = 0, f (2) = 1, thus, f has a maximum of f (x) does not exist at x = 3/2. f (â&#x2C6;&#x2019;2) = 7, f 2 7 at x = â&#x2C6;&#x2019;2 and a minimum of 0 at x = 3/2. Ď&#x20AC; 2 Absolute minimum of 0 at 0
4.5.19 Absolute maximum of 4 at
4.5.20 Absolute maximum of 16 at 2 Absolute minimum of 0 at 0
4.5.21 Absolute maximum of 8.43 at 1.78 Absolute minimum of 0 at 0
4.5.22 Absolute maximum of 3.56 at 0.86 Absolute minimum of 0 at 0