Pre-Calc Homework Solutions 289

Pre-Calc Homework Solutions 289 - Section 7.1 Quick Review...

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Unformatted text preview: Section 7.1 Quick Review 7.1 1. On the interval, sin 2x 0 when x 2 x x2 1 289 7. , 0, or 2 0 when x 1, x x2 1 0. Test one point on each subinterval: 1 ; for x 2 . Test one 1; for x 1, x x2 1 point on each subinterval: for x for x x and 4 3 , sin 2x 4 4 1 . The 2 function changes sign at 0. The graph is + f (x) x 30 , sin 2x 1; for x , sin 2x 1; and for 2 3 , sin 2x 4 2 1. The function changes sign at , 0, x2 x2 5 0 . The graph is 8. + 0 2 2 + 2 f (x) x 2 4 0 when x 2 and is undefined when 5 , 2 3 x x2 x2 x2 x2 x2 x2 2 4 2 4 2 4 2. Test one point on each subinterval: for x 2. x 2 3x 2 (x 1)(x 2) 0 when x 0, x 2 3, 1 or 2. Test 3x 2 2; one point on each subinterval: for x for x x 2 3 2 ,x 2 3x 2 1 ; and for x 4 17 x2 2 ; for x 1.9, 2 4.13; for x 0, 9 x 4 2 1 x 2 5 ; for x 1.9, 2 4.13; and for x , 2 x 4 2 17 . The function changes sign at 2, 2, 2 9 and 2. The graph is + + 2 2 + 3 f (x) x 3x 2 2. The function changes sign at 1 and 2. The graph is + 2 1 2 + 4 f (x) x 3 2 2 9. sec (1 x x 0.9633 1 sin2 x) 1 cos (1 cos x ) is undefined when 3. x 2 2x 3 0 has no real solutions, since b 2 4ac ( 2)2 4(1)(3) 8 0. The function is always positive. The graph is + 4 2 f (x) x k or 2.1783 1 sin2 0) 1 k for any integer k. Test for 2.4030. sin2 1) 32.7984. The 0: sec (1 Test for x 1: sec (1 sign alternates over successive subintervals. The function 1) 0 when x 1, 1 1; and 1 or 1. 2 4. 2x 3 3x 2 1 (x 1)2(2x changes sign at 0.9633 integer. The graph is + k or 2.1783 k , where k is an Test one point on each subinterval: for x 2x 3 x 3x 2 3 , 2x 3 2 1 3x 2 4; for x 1 0, 2x 3 3x 2 + f (x) x 1. The function changes sign at 2.1783 0.9633 0.9633 2.1783 1 . The graph is 2 2 1 2 + 1 + 2 f (x) x 10. On the interval, sin 1 x 0 when x point on each subinterval: for x x 3 5 0, , , or . 4 4 4 8 2 1 1 or . Test one 3 2 1 0.1, sin 0.54; for x 0.15, sin 1 x 1 x 0.37; and for x 0.2, 1 1 , and . 3 2 sin 0.96. The graph changes sign at 5. On the interval, x cos 2x 0 when x The graph is 0.1 1 3 Test one point on each subinterval: for x x cos 2x x cos 2x 2 16 , , + 1 2 ; for x 2 , x cos 2x ; for x f (x) x 0.2 ; and for x 4 4, x cos 2x , 0.58. The function changes sign at + 0 4 3 5 , and . The graph is 4 4 f (x) x 3 4 + 5 4 4 6. xe 0 when x always positive. x 0. On the rest of the interval, xe x is ...
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