Find the domain of the logarithmic function...
f(x)= log (x^2 - 6x + 8) I started doing this and then kinda got stuck, can someone help me...?
f(x)= log (x^2 - 6x + 8) I started doing this and then kinda got stuck, can someone help me...?
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The log function is not defined for non-positive arguments. Therefore the domain of the function is:
x² - 6x + 8 > 0
x² - 6x > -8
x² - 6x + 9 > 1
(x-3)² > 1
x-3 > 1
x-3 < -1
x > 4
x < 2
The domain of the function is: (-inf, 2) U (4, inf) or all x except 2 ≤ x ≤ 4
Plot the function f(x) = x² - 6x + 9 and you will see that it's a parabola with x-intercepts at 2 and 4, is negative between those intercepts and positive everywhere else.
x² - 6x + 8 > 0
x² - 6x > -8
x² - 6x + 9 > 1
(x-3)² > 1
x-3 > 1
x-3 < -1
x > 4
x < 2
The domain of the function is: (-inf, 2) U (4, inf) or all x except 2 ≤ x ≤ 4
Plot the function f(x) = x² - 6x + 9 and you will see that it's a parabola with x-intercepts at 2 and 4, is negative between those intercepts and positive everywhere else.
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logarithm is defined only if the argument (whatever thats inside log) is positive.
So, f(x)= log (x² - 6x + 8) is defined for x such that x² - 6x + 8 > 0
x² - 6x + 8 > 0
x² - 4x - 2x + 8 > 0
(x - 4) (x - 2) > 0
Therefore, our domain is, x ϵ (-∞ , 2) U (4, ∞)
or for every x > 4 or x < 2
So, f(x)= log (x² - 6x + 8) is defined for x such that x² - 6x + 8 > 0
x² - 6x + 8 > 0
x² - 4x - 2x + 8 > 0
(x - 4) (x - 2) > 0
Therefore, our domain is, x ϵ (-∞ , 2) U (4, ∞)
or for every x > 4 or x < 2
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f(x) = log(x² - 6x + 8)
x² - 6x + 8 must be positive.
Use the quadratic formula to solve x² - 6x + 8 = 0: x = 2, 4.
The domain is (0, 2) ∪ (2, 4) ∪ (4, +∞)
x² - 6x + 8 must be positive.
Use the quadratic formula to solve x² - 6x + 8 = 0: x = 2, 4.
The domain is (0, 2) ∪ (2, 4) ∪ (4, +∞)
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yes.
(x-4)(x+2) >0
so x must be greater than four, or less than negative 2. domain: (-<><>,-2) U (4, <><> )
(x-4)(x+2) >0
so x must be greater than four, or less than negative 2. domain: (-<><>,-2) U (4, <><> )