Help with theses functions?
- Thread starter Zfuss12
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Daniel_Feldman
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I'll show you domains, and see if you can work from there to find the ranges.
f(x)= (3x-80/RADICAL (15-6x)
Remember that the quantity under the radical must be non-negative. Thus, the domain would be 15-6x greater than or equal to 0, so x is less than or equal to 15/6, right?
Wrong! Since the radical is the denominator, it cannot equal 0. So the domain is x less than 15/6.
f(x)=7x+12
No restrictions here-the domain is all real numbers.
f(x)= RADICAL(5x-1)/(8x-16)
First off, 5x-1 must be greater than or equal to 0, so x must be greater than or equal to 1/5. Also, 8x-16 cannot equal 0, so x cannot equal 2.
So the domain is [1/5, 2) U (2, infinity)
f(x)= (3x-80/RADICAL (15-6x)
Remember that the quantity under the radical must be non-negative. Thus, the domain would be 15-6x greater than or equal to 0, so x is less than or equal to 15/6, right?
Wrong! Since the radical is the denominator, it cannot equal 0. So the domain is x less than 15/6.
f(x)=7x+12
No restrictions here-the domain is all real numbers.
f(x)= RADICAL(5x-1)/(8x-16)
First off, 5x-1 must be greater than or equal to 0, so x must be greater than or equal to 1/5. Also, 8x-16 cannot equal 0, so x cannot equal 2.
So the domain is [1/5, 2) U (2, infinity)
mmm4444bot
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Reduce the Fraction
Hello Zfuss12:
I agree with Daniel_Feldman's solutions; I would only add that the fraction 15/6 in the first problem can be reduced since the numerator and denominator share 3 as a common factor.
In other words: x < 5/2.
~ Mark
Hello Zfuss12:
I agree with Daniel_Feldman's solutions; I would only add that the fraction 15/6 in the first problem can be reduced since the numerator and denominator share 3 as a common factor.
In other words: x < 5/2.
~ Mark
I guessed where some () were. If I missed that could change things. Also I got interuppted so I'm in third place. Sorry Gang.
Whenever you see a radical, check for negatives and eliminate them.
Eliminate division by zero.
f(x)= (3x-80)/RADICAL (15-6x)
If x > 2.5 it is out
If x = 2.5 it is out
If x < 2.5 but close, f(x) -> -oo
Any negative x is ok.
The domain is -oo < x < 2.5
The range is -oo < y < ?
I don't see an algebraic way to find the ?. Other methods (calculus, computer, trial-and-error, etc.) give about (-65/3,-12) as the max.
f(x)=7x+12
An easy one. In a linear equation nothing is eliminated. Domain and range are -oo to +oo
f(x)= RADICAL(5x-1)/(8x-16)
x > .2
x not = 2
f(x)=-oo to +oo 'cause near x=2 it gets as big as you want.
Whenever you see a radical, check for negatives and eliminate them.
Eliminate division by zero.
f(x)= (3x-80)/RADICAL (15-6x)
If x > 2.5 it is out
If x = 2.5 it is out
If x < 2.5 but close, f(x) -> -oo
Any negative x is ok.
The domain is -oo < x < 2.5
The range is -oo < y < ?
I don't see an algebraic way to find the ?. Other methods (calculus, computer, trial-and-error, etc.) give about (-65/3,-12) as the max.
f(x)=7x+12
An easy one. In a linear equation nothing is eliminated. Domain and range are -oo to +oo
f(x)= RADICAL(5x-1)/(8x-16)
x > .2
x not = 2
f(x)=-oo to +oo 'cause near x=2 it gets as big as you want.