# Student Study and Solutions Manual for Larson's Precalculus with Limits, 3rd

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Area: ( −∞, − 1] ∪ [1, ∞) (a) tan θ = ⎡ π ⎞ ⎛ π⎤ diversity: ⎢− , zero ⎟ ∪ ⎜ zero, ⎥ ⎣ 2 ⎠ ⎝ 2⎦ 20 forty-one ⎛ 20 ⎞ θ = arctan ⎜ ⎟ ≈ 26. zero° ⎝ forty-one ⎠ h 50 h = 50 tan 26° ≈ 24. four ft (b) tan 26° = 109. (a) tan θ = x 20 θ = arctan 119. y = arcsec x 20 zero ≤ y < five ≈ 14. zero° 20 12 x = 12: θ = arctan ≈ 31. zero° 20 (b) x = five: θ = arctan 1 sin x − − The functionality sin −1 x is similar to arcsin x, that is 1 is the sin x reciprocal of the sine functionality and is comparable to csc x. area: ( −∞, ∞) diversity: (0, π ) 2 < y ≤ π ⇒ y = π four threeπ four π 2 ≤ y < 0∪0 < y ≤ π 2 ⇒ y = π 6 ⎛2 three⎞ 2 three one hundred twenty five.

H( − x ) = f ( − x ) ± g ( − x ) = − f ( x) ± g ( x) simply because f is bizarre and g is even ≠ h( x ) ≠ − h( x ) So, h( x) is neither abnormal nor even. five. f ( x) = a2 n x 2 n + a2 n − 2 x 2 n − 2 + " + a2 x 2 + a0 f ( − x) = a2 n ( − x) 2n + a2 n − 2 ( − x) 2n − 2 + " + a2 (− x) + a0 = a2 n x 2 n + a2 n − 2 x 2 n − 2 + " + a 2 x 2 + a0 = f ( x) 2 So, f ( x) is even. 7. (a) April eleven: 10 hours (b) pace = April 12: 24 hours distance 2100 a hundred and eighty five = = = 25 mph 2 time 7 7 eighty one three April thirteen: 24 hours 2 April 14: 23 hours three overall: 2 eighty one hours three one hundred eighty t + 3400 7 1190 area: zero ≤ t ≤ nine diversity: zero ≤ D ≤ 3400 (c) D = − (d) Copyright 2013 Cengage studying.

585 ln 2 sixty one. (a) log1 2 five = area: (0, ∞) (b) log1 2 five = ln x + three = zero ln x = −3 x-intercept: (e −3 , zero) = log 2 + 2 log three ≈ 1. 255 Vertical asymptote: x = zero f ( x) ln five ≈ −2. 322 ln (1 2) sixty three. log 18 = log( 2 ⋅ 32 ) x = e −3 x log five ≈ −2. 322 log(1 2) 1 2 three 1 2 1 four three three. sixty nine four. 10 2. 31 1. sixty one sixty five. ln 20 = ln ( 22 ⋅ five) = 2 ln 2 + ln five ≈ 2. 996 sixty seven. log five five x 2 = log five five + log five x 2 = 1 + 2 log five x sixty nine. log three nine = log three nine − log three x x = log three 32 − log three x1 2 = 2− 1 log three x 2 Copyright 2013 Cengage studying.

The graph rises to the left and rises to the ideal. four three 2 (−1, zero) 1 x −4 −3 −2 1 2 three four −3 −4 21. (a) f ( x) = 3x3 − x 2 + three x f ( x) –3 –2 −1 zero 1 2 three – 87 – 25 –1 three five 23 seventy five The 0 is within the period [−1, 0]. (b) 0: x ≈ −0. 900 Copyright 2013 Cengage studying. All Rights Reserved. will not be copied, scanned, or duplicated, in entire or partially. 116 bankruptcy 2 Polynomial and Rational capabilities 6x + three 23. five x − three 30 x 2 − 3x + eight 29. eight + 31. (7 + 5i ) + ( − four + 2i ) = (7 − four) + (5i + 2i ) = three + 7i 30 x 2 − 18 x 15 x + eight 33.

F ( x ) = ( x − 2)( x + 6) sixty seven. f ( x) = ( x − 0)( x + 5)( x − 1) = x 2 + four x − 12 observe: f ( x) = a( x − 2)( x + 6) has zeros 2 and − 6 for = x( x 2 + four x − five) all genuine numbers a ≠ zero. = x3 + four x 2 − five x notice: f ( x) = ax( x 2 + four x − 5), a ≠ zero, has measure three fifty nine. f ( x) = ( x − 0)( x + 4)( x + five) and zeros x = zero, − five, and 1. = x( x 2 + nine x + 20) = x3 + nine x 2 + 20 x notice: f ( x) = ax( x + 4)( x + five) has zeros zero, − four, and ( sixty nine. f ( x ) = ( x − zero) x − −5 for all genuine numbers a ≠ zero. ( = x x − )( ( three x − − three )( three x + ) )) three = x3 − 3x observe: f ( x) = a( x3 − 3x), a ≠ zero, has measure three and zeros x = zero, three, and − three.