SOLUTION: Can you please help me with this problem: Find each value given that f(x)=2x+1
f(3)+f(2).
And: Find the domain and range of f(x)= 1/x-4. I know the domain is x and the range is
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Question 16555: Can you please help me with this problem: Find each value given that f(x)=2x+1
f(3)+f(2).
And: Find the domain and range of f(x)= 1/x-4. I know the domain is x and the range is y, but how do I figure it on a fraction?
Thank You
Found 2 solutions by venugopalramana, rapaljer:
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
f(x)=2x+1
f(3)+f(2).
to get f(3) you have to substitute 3 inplace of x in the given function .hence
f(3)=2*3+1=7
f(2)=2*2+1=5
f(3)+f(2)=7+5=12
Find the domain and range of f(x)= 1/x-4. I know the domain is x and the range is y, but how do I figure it on a fraction?
domain of a function means the value x can take to make the function meaningfull or legal mathematically...for example in maths division by zero has no meaning.similarly square root of a negative number does not exist.
here we have 1/(x-4)..which has a division ..so x-4 cannot be zero..so x cannot be equal to 4...so the domain of x is all values of real numbers except 4
now coming to range it is those values of y ,that are obtained by taking all possible value of x in its domain as determined above.
here we find that y can take all values of real numbers since x will not equal 4 in the domain.so range of y is all possible values of real numbers
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
I need to add a note to the previous solution that was posted to this problem. The range of f(x)= 1/(x-4) is NOT all real numbers. If you draw the graph of y=1/(x-4) you will notice that the value of y can approach zero, but it will NEVER equal ZERO! So the range turns out to be all real values of y NOT EQUAL TO ZERO!
Consider the graph:
R^2 at SCC
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