1.14
INEQUALITIES
left of x = -4, both x — 1 and x + 4 are negative and, therefore, g(x) is positive. As we pass through
x = — 4, jr + 4 changes sign and g(x) becomes negative. When we pass through * = 1, A: - 1 changes sign
and g(x) becomes and then remains positive. Thus, (x - \)(x + 4) is negative for -4 < x < 1. Answer
Fig. 1-4
1.15
1.16
Fig. 1-5
Solve x
2 - 6x + 5 > 0.
both .* - 1 and jc - 5 are negative and, therefore, h(x) is positive. When we pass through x = \, x-\
changes sign and h(x) becomes negative. When we move further to the right and pass through x = 5, x — 5
changes sign and h(x) becomes positive again. Thus, h(x) is positive for x < 1 and for x>5.
Answer x > 5 or x < 1. This is the union of the intervals (5, °°) and (—°°, 1).
Solve x2 + Ix - 8 < 0.
are negative and, therefore, F(x) = (x + 8)(x - 1) is positive. When we pass through x = -8, x + 8
changes sign and, therefore, so does F(x). But when we later pass through x = l, x-l changes sign and
F(x) changes back to being positive. Thus, F(x) is negative for -8 < x < 1. Answer
Fig. 1-6
1.17
1.18
1.19
Fig. 1-7
Solve 5x - 2x
2 > 0.
are x = 0 and *=|. For x
5 — 2x changes sign and, therefore, G(x) changes back to being negative. Thus, G(x) is positive when and only
when 0 < x < |. Answer
Solve (Jt-l)
2 (* + 4)<0.
* + 4<0 and jc^l.
Answer x<— 4 [In interval notation, (—=°, — 4).]
the left of — 1, x, x — 1, and x + 1 all are negative and, therefore, H(x) is negative. As we pass
through x = — 1, x + 1 changes sign and, therefore, so does H(x). When we later pass through x = 0, x
changes sign and, therefore, H(x) becomes negative again. Finally, when we pass through x = l, x-\
changes sign and H(x) becomes and remains positive. Therefore, H(x) is positive when and only when
— 1 < A: < 0 or x>\. Answer
Solve (x-l)(x + 4)<0.
Solve x(x-l)(x + l)>0.
3
The key points of the function g(x) = (x - l)(x + 4) are x = — 4 and x = l (see Fig. 1-4). To the
Factor: x2 -6x + 5 = (x - l)(x - 5). Let h(x) = (x - \)(x - 5). To the left of x = 1 (see Fig. 1-5),
Factor: x2 + Ix - 8 = (x + &)(x - 1), and refer to Fig. 1-6. For jc<-8, both x + 8 and x-l
Factor: 5x - 2x2 = x(5 - 2x), and refer to Fig. 1-7. The key points for the function G(x) = x(5 - 2x)
(x — I)2 is always positive except when x = 1 (when it is 0). So, the only solutions occur when
The key points for H(x) = x(x - l)(x + 1) are x = 0, x = l, and jc=-l (see Fig. 1-8). For x to
INEQUALITIES
left of x = -4, both x — 1 and x + 4 are negative and, therefore, g(x) is positive. As we pass through
x = — 4, jr + 4 changes sign and g(x) becomes negative. When we pass through * = 1, A: - 1 changes sign
and g(x) becomes and then remains positive. Thus, (x - \)(x + 4) is negative for -4 < x < 1. Answer
Fig. 1-4
1.15
1.16
Fig. 1-5
Solve x
2 - 6x + 5 > 0.
both .* - 1 and jc - 5 are negative and, therefore, h(x) is positive. When we pass through x = \, x-\
changes sign and h(x) becomes negative. When we move further to the right and pass through x = 5, x — 5
changes sign and h(x) becomes positive again. Thus, h(x) is positive for x < 1 and for x>5.
Answer x > 5 or x < 1. This is the union of the intervals (5, °°) and (—°°, 1).
Solve x2 + Ix - 8 < 0.
are negative and, therefore, F(x) = (x + 8)(x - 1) is positive. When we pass through x = -8, x + 8
changes sign and, therefore, so does F(x). But when we later pass through x = l, x-l changes sign and
F(x) changes back to being positive. Thus, F(x) is negative for -8 < x < 1. Answer
Fig. 1-6
1.17
1.18
1.19
Fig. 1-7
Solve 5x - 2x
2 > 0.
are x = 0 and *=|. For x
5 — 2x changes sign and, therefore, G(x) changes back to being negative. Thus, G(x) is positive when and only
when 0 < x < |. Answer
Solve (Jt-l)
2 (* + 4)<0.
* + 4<0 and jc^l.
Answer x<— 4 [In interval notation, (—=°, — 4).]
the left of — 1, x, x — 1, and x + 1 all are negative and, therefore, H(x) is negative. As we pass
through x = — 1, x + 1 changes sign and, therefore, so does H(x). When we later pass through x = 0, x
changes sign and, therefore, H(x) becomes negative again. Finally, when we pass through x = l, x-\
changes sign and H(x) becomes and remains positive. Therefore, H(x) is positive when and only when
— 1 < A: < 0 or x>\. Answer
Solve (x-l)(x + 4)<0.
Solve x(x-l)(x + l)>0.
3
The key points of the function g(x) = (x - l)(x + 4) are x = — 4 and x = l (see Fig. 1-4). To the
Factor: x2 -6x + 5 = (x - l)(x - 5). Let h(x) = (x - \)(x - 5). To the left of x = 1 (see Fig. 1-5),
Factor: x2 + Ix - 8 = (x + &)(x - 1), and refer to Fig. 1-6. For jc<-8, both x + 8 and x-l
Factor: 5x - 2x2 = x(5 - 2x), and refer to Fig. 1-7. The key points for the function G(x) = x(5 - 2x)
(x — I)2 is always positive except when x = 1 (when it is 0). So, the only solutions occur when
The key points for H(x) = x(x - l)(x + 1) are x = 0, x = l, and jc=-l (see Fig. 1-8). For x to