212. I need to find the domain of f(x) = log(3x-x^2).... I know it would be the inverse of 10^(3x-x^2).... I think? Ah! Help, explain how I would go about figuring this out... thanks
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You can only take the log of positive numbers, so
3x-x^2 > 0
x(3-x) > 0
If the product of two numbers is positive then both numbers are positive or both numbers are negative. Examine both case.
So, x> 0 and (3-x) > 0 OR x<0 and (3-x) < 0
x > 0 and x < 3 OR x < 0 and x > 3. The second case is empty.
The domain is 0 < x < 3.
You can only take the log of positive numbers, so
3x-x^2 > 0
x(3-x) > 0
If the product of two numbers is positive then both numbers are positive or both numbers are negative. Examine both case.
So, x> 0 and (3-x) > 0 OR x<0 and (3-x) < 0
x > 0 and x < 3 OR x < 0 and x > 3. The second case is empty.
The domain is 0 < x < 3.
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3x - x^2 must always be greater than zero, because the log of a negative number is undefined.
3x - x^2 > 0
x(3 - x) > 0
zeros: x = 0; x = 3
Plot the zeros on a number line and identify the intervals:
....(-∞, 0)..|......(0, 3)......|.....(3, ∞).....
<------------+-------------------+-----…
...............0...................3
Intervals:
(-∞, 0)
(0, 3)
(3, ∞)
Identify a test number in each interval and then test x(3 - x) at each interval:
(-∞, 0): -1; -1(3 - -1) = -1(3 + 1) = -4 < 0; fail
(0, 3): 1; 1(3 - 1) = 1(2) = +2 > 0; pass
(3, ∞): 4; 4(3 - 4) = 4(-1) = -4 < 0; fail
Answer:
Domain of f(x) = log(3x-x^2): (0, 3) or 0 < x < 3
3x - x^2 > 0
x(3 - x) > 0
zeros: x = 0; x = 3
Plot the zeros on a number line and identify the intervals:
....(-∞, 0)..|......(0, 3)......|.....(3, ∞).....
<------------+-------------------+-----…
...............0...................3
Intervals:
(-∞, 0)
(0, 3)
(3, ∞)
Identify a test number in each interval and then test x(3 - x) at each interval:
(-∞, 0): -1; -1(3 - -1) = -1(3 + 1) = -4 < 0; fail
(0, 3): 1; 1(3 - 1) = 1(2) = +2 > 0; pass
(3, ∞): 4; 4(3 - 4) = 4(-1) = -4 < 0; fail
Answer:
Domain of f(x) = log(3x-x^2): (0, 3) or 0 < x < 3