Question 218207: Can someone help me with this please. IF f(x)=3x+1, and g(x)=(x^*2+5x)^*-1/2, find g(f(x)). Any explanation on how to do it would be appreciated..
Found 3 solutions by stanbon, Alan3354, Theo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Can someone help me with this please. IF f(x)=3x+1, and g(x)=(x^*2+5x)^*-1/2, find g(f(x)).
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g[f(x)] = g[3x+1]
= [(3x+1)^2+5x)^(-1/2)]
= [(3x^2+6x+1+5x)^(-1/2)]
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= 1/sqrt(3x^2+11x+1)
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Cheers,
Stan H.
Answer by Alan3354(69443)  (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! This would look like the following if my assumption about what you are writing are correct:
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Whether my assumptions are correct or not, the concept is the same and you can apply it to the correct formula if I got it wrong.
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the function is a set of rules that operate on the argument to get a result.
If you say f(x) = x^2, then the function name is f and the argument is x and the set of rules are to square the argument to get a result.
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If you let x = 3, then f(x) = x^2 becomes f(3) = (3)^2
the name of the function is the same,
the argument changes from x to 3.
the set of rules is the same and operates on the argument which in this case is 3 rather than x.
the only thing you changed is the argument. everything else remains the same.
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now change the argument to (x^3)
f(x) = x^2 becomes f(x^3) = (x^3)^2 = x^6
the name of the function is the same.
the set of rules is the same.
the argument has changed so the same set of rules is being applied to the new argument of (x^3) rather than x.
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this is exactly what you are doing when you take g(f(x))
the argument that was x is being replaced with f(x)
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your equations are:
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f(x)=3x+1
and:
g(x)=(x^2+5x)^(-1/2)
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your first function is f(x) = 3x+1.
the argument is x and the function being performed is to take 3 times it and add 1 to it.
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your second function is g(x) = (x^2 + 5x)^(-1/2)
the argument is x and the function being performed is to square it and then add 5 times it and then raise the sum of them to the (-1/2) power.
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when you take g(f(x)), then the argument becomes f(x) rather than x. Since f(x) = (3x+1), then the new argument is (3x+1) and the function being performed remains the same except it is operating on (3x+1) rather than x.
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the function g(x) operating on the argument x is:
g(x) = (x^2 + 5x)^(-1/2)
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the function g(x) operating on the argument (3x+1) is:
g(f(x)) = g(3x+1) = ((3x+1)^2 + 5*(3x+1))^(-1/2)
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These formulas look like the following after putting them through the formula generating functions of algebra.com:
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This is your answer and should be correct, but you can simplify it further if you wish.
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This means to remove parentheses and combine like terms and take the negative out of the exponent I believe.
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We'll do that and see what we get.
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your equation of:
becomes:
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How do we know we did this right?
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take any value of x and place it in the original equations and then place it in the final equation and see if you get the same answer.
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let's take x = 7
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when x = 7, then:
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when x = 22, then:
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when x = 7, then:
this becomes:
this becomes:
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The answers come out the same so the formulas are equivalent.
g(x) when x = 22 is the same as g(f(x)) when x = 7.
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Your answer is:
simplified as far as I can take it.
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That same answer is:
before the simplification process was done.
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