Domain and Range of Functions

• Workshop solutions to Sections 2.1 and 2.2(1.1 & 1.2) have been provided

Domain and Range of Functions

• To find the domain of the function f(x) = 9 - x², note that since f(x) is a polynomial, the domain Df = R = (-∞, ∞)
• The domain of any polynomial is R
• To find the range of the function f(x) = 9 − x², Rf = (-∞, 9]
• To find the domain of the function f(x) = 6 - 2x, since f(x) is a polynomial, then Df = R = (-∞, ∞)
• To find the range of the function f(x) = 6 – 2x, since f(x) is a polynomial of degree one (i. e. is of an odd degree), then Rf = R = (-∞, ∞)
• To find the domain of the function f(x) = x² – 2x – 3, since f(x) is a polynomial, then Df = R = (-∞, ∞)
• To find the domain of the function f(x) = 1 + 2x³ – x⁵, since f(x) is a polynomial, then Df = R = (-∞, ∞)
• To find the domain of the function f(x) = 5, since f(x) is a polynomial, then Df = R = (-∞, ∞)
• To find the range of the function f(x) = 5, then Rf = {5}
• To find the domain of the function f(x) = |x − 1|, since f(x) is an absolute value of a polynomial, then Df = R = (-∞, ∞)
• To find the domain of the function f(x) = |x + 5|, since f(x) is an absolute value of a polynomial, then Df = R = (-∞, ∞)
• The domain of an absolute value of any polynomial is R
• To find the domain of the function f(x) = |x|, since f(x) is an absolute value of a polynomial, then Df = R = (-∞, ∞)
• To find the range of the function f(x) = |x|, Rf = [0,∞)
• The range of an absolute value of any polynomial is always [0,∞)
• To find the domain of the function f(x) = |3x − 6|, since f(x) is an absolute value of a polynomial, then Df = R = (-∞, ∞)
• To find the domain of the function f(x) = |9 – 3x|, since f(x) is an absolute value of a polynomial, then Df = R = (-∞, ∞)
• To find the domain of the function f(x) = (x-2)/(x+3), f(x) is defined when x + 3 ≠ 0 ⇒ x = −3, so Df = R \ {-3} = (-∞, −3)U(−3,∞)
• To find the domain of the function f(x) = (x+2)/(x-3), f(x) is defined when x − 3 ≠ 0 ⇒ x ≠ 3, so Df = R \ {3} = (-∞, 3)U(3,∞)
• To find the domain of the function f(x) = (x+2)/(x²-9), f(x) is defined when x² − 9 ≠ 0 ⇒ x² + 9 ⇒ x ≠ ±3, so Df = R \ {−3,3} = (-∞, −3)U(−3,3)U(3,∞)
• To find the domain of the function f(x) = (x+2)/(x² - 5x + 6), f(x) is defined when x² – 5x + 6 ≠ 0 ⇒ (x - 2)(x - 3) ≠ 0 ⇒ x ≠ 2 or x ≠ 3, so Df = R \ {2,3} = (−∞, 2)U(2,3)U(3,∞)
• To find the domain of the function f(x) = (x+2)/(x²-x-6), f(x) is defined when x² − x − 6 ≠ 0 ⇒ (x + 2)(x − 3) ≠ 0 ⇒ x = −2 or x ≠ 3, so Df = R \ {−2,3} = (-∞, −2)U(−2,3)U(3,∞)
• To find the domain of the function f(x) = (x+2)/(x²+9), f(x) is defined when x² + 9 ≠ 0 but for any value x the denominator x² + 9 cannot be 0, so Df = R = (-∞, ∞)
• To find the domain of the function f(x) = ³√(x - 3), Df = R = (-∞, ∞)
• The domain of an odd root of any polynomial is R
• To find the domain of the function f(x) = √(x - 3), f(x) is defined when x − 3 ≥ 0 ⇒ x ≥ 3 because f(x) is an even root, so Df = [3,∞)
• To find the domain of the function f(x) = √(3 - x), f(x) is defined when 3 − x ≥ 0 ⇒ −x ≥ −3 ⇒ x ≤ 3 because f(x) is an even root, so Df = (-∞, 3]
• To find the domain of the function f(x) = √(x + 3), f(x) is defined when x + 3 ≥ 0 ⇒ x ≥ −3 because f(x) is an even root, so Df = [−3,∞)
• To find the domain of the function f(x) = √(-x), f(x) is defined when -x ≥ 0 ⇒ x ≤ 0 because f(x) is an even root, so Df = (-∞, 0]
• To find the range of the function f(x) = √(-x), then Rf = [0,∞)
• The range of an even root is always ≥ 0
• To find the domain of the function f(x) = √(9-x²), f(x) is defined when 9 − x² ≥ 0 ⇒ −x² ≥ −9 ⇒ x² ≤ 9 ⇒ √x² < √9 = |x| ≤ 3 ⇒ −3 ≤ x ≤ 3, so Df = [-3,3]
• To find the domain of the function f(x) = (x+2)/(√(x - 3)), f(x) is defined when x − 3 > 0 ⇒ x>3, so Df = (3,∞)
• To find the domain of the function f(x) = (x+2)/(√(9-x²)), f(x) is defined when 9 − x² > 0 ⇒ −x² > −9 ⇒ x² < 9 ⇒ √x2 < √9 ⇒ |x| < 3 ⇒ -3 < x <3, so Df = (-3,3)
• To find the domain of the function f(x) = √(x² - 9), f(x) is defined when x² − 9 ≥ 0 ⇒ x² ≥ 9 ⇒ √x² ≥ √9 = |x| ≥ 3 ⇒ x ≥ 3 or x ≤ −3, so Df = (-∞, −3] U [3,∞)
• To find the range of the function f(x) = √(x²- 9), then Rf = [0, ∞)
• To find the domain of the function f(x) = (x+2)/(√(x²-9)), f(x) is defined when x² − 9 > 0 ⇒ x² > 9 ⇒ √x² > √9 = |x| > 3 ⇒ x > 3 or x <-3, so Df = (-∞, −3) ∪ (3,∞)
• To find the domain of the function f(x) = √(9 + x²), f(x) is defined when 9 + x² ≥ 0 but it is always true for any value x, so Df = R
• To find the domain of the function f(x) = √(x² - 25), f(x) is defined when x² – 25 ≥ 0 ⇒ x² ≥ 25 ⇒ √x² ≥ √25 = |x| ≥ 5 ⇒ x ≥ 5 or x ≤ −5, so Df = (-∞, −5] U [5, ∞)
• To find the domain of the function f(x) = √(16 - x²), f(x) is defined when 16 − x² ≥ 0 ⇒ −x² ≥ −16 ⇒ x² < 16 ⇒ √x² < √16 ⇒ |x| ≤ 4 ⇒ −4 ≤ x ≤ 4, so Df = [−4,4]
• To find the range of the function f(x) = √(16 - x²), we know that f(x) is defined when 16 − x² ≥ 0 ⇒ −x² ≥ −16 ⇒ x² ≤ 16 ⇒ √x² < √16 ⇒ |x| ≤ 4 ⇒ −4 ≤ x ≤ 4, so Df = [−4,4]. Using Df we find the outputs vary from 0 to 4, hence, Rf = [0,4]
• To find the domain of the function f(x) = (x + |x|)/x, f(x) is defined when x ≠ 0, so Df = R \ {0} = (-∞, 0) ∪ (0, ∞)
• To find the domain of the function f(x) = {1, x<0 ; x, x ≥ 0}, it is clear from the definition of the function f(x) that Df = R = (-∞, ∞)
• To find the domain of the function f(x) = (2 - √x)/(√(x² + 1)), D√x = [0,∞), f(x) is defined when - x ≥ 0. x² + 1 > 0 but this is always true for all x ⇒ D√x²+1 = R. Hence, Df = D√x ∩ D√x²+1 = [0, ∞) ∩ R = [0,∞)
• To find the domain of the function f(x) = √(x − 1) + √(x + 3), f(x) is defined when x − 1 ≥ 0 ⇒ x ≥ 1 ⇒ D√(x-1) = [1, ∞), x + 3 ≥ 0 ⇒ x ≥ −3 ⇒ D√(x+3) = [−3, ∞). Hence, Df = D√(x-1) ∩ D√(x+3) = [1, ∞) n [−3, 8) = [1, 8)
• The function f(x) = 3x⁴ + x² + 1 is a polynomial function
• The function f(x) = 5x³ + x² + 7 is a cubic function
• The function f(x) = −3x² + 7 is a quadratic function
• The function f(x) = 2x + 3 is a linear function
• The function f(x) = x⁷ is a power function
• The function f(x) = (2x+3)/(x²-1) is a rational function
• The function f(x) = (x-3)/(x+2) is a rational function and it is an algebraic function as well
• The function f(x) = sin x is a trigonometric function
• The function f(x) = eˣ is a natural exponential function
• The function f(x) = 3ˣ is a general exponential function
• The function f(x) = x² + √x - 2 is an algebraic function
• The function f(x) = −3 is a constant function
• The function f(x) = log₃x is a general logarithmic function
• The function f(x) = ln x is a natural logarithmic function
• For the function f(x) = 3x⁴ + x² + 1, f(x) is an even function
• For the function f(x) = 9 − x², f(x) is an even function
• For the function f(x) = x⁵ − x, f(x) is an odd function
• For the function f(x) = 2 – √x, f(x) is neither even nor odd
• For the function f(x) = 3x + 2/√(x²+9), f(x) is neither even nor odd
• For the function f(x) = 3/√(x²+9), f(x) is an even function
• For the function f(x) = √(4 + x²), f(x) is an even function
• For the function f(x) = 3, f(x) is an even function
• For the function f(x) = (9-x²)/(x-2), f(x) is neither even nor odd
• For the function f(x) = (x²-4)/(x²+1), f(x) is an even function
• For the function f(x) = 3|x|, f(x) is an even function
• For the function f(x) = x⁻², f(x) is an even function
• For the function f(x) = x³ – 2x + 5, f(x) is neither even nor odd
• For the function f(x) = √(x⁵ − x³ + x), f(x) is an odd function
• For the function f(x) = 7, f(x) is an even function
• For the function f(x) = (x³-4)/(x³+1), f(x) is neither even nor odd
• For the function f(x) = (x²-1)/(x³+3), f(x) is neither even nor odd
• For the function f(x) = x⁶ – 4x² + 1, f(x) is an even function
• The function f(x) = x² is increasing on (0, ∞)
• The function f(x) = x² is decreasing on (18,0)
• The function f(x) = x³ is increasing on (18,8)
• The function f(x) = x³ is not decreasing at all
• The function f(x) = √x is increasing on (0,∞)
• The function f(x) = √x is not decreasing at all
• The function f(x) = 1/x is not increasing at all
• The function f(x) = 1/x is decreasing on (-∞, ∞).{0}