SOLUTION: f(x)= "3x-12"-under square root symbol. find the specified function value, if it exists. f(9)= f(4)= f(1)= f(-1)=

Algebra ->  Equations -> SOLUTION: f(x)= "3x-12"-under square root symbol. find the specified function value, if it exists. f(9)= f(4)= f(1)= f(-1)=      Log On


   

Question 937940: f(x)= "3x-12"-under square root symbol.
find the specified function value, if it exists.
f(9)=
f(4)=
f(1)=
f(-1)=
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29+=+sqrt%283x-12%29
f%289%29+=+sqrt%283%289%29-12%29
f%289%29+=+sqrt%2827-12%29
f%289%29+=+sqrt%2815%29 Exact value
f%289%29+=+3.872983 Approximate value
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f%28x%29+=+sqrt%283x-12%29
f%284%29+=+sqrt%283%284%29-12%29
f%284%29+=+sqrt%2812-12%29
f%284%29+=+sqrt%280%29
f%284%29+=+0
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f%28x%29+=+sqrt%283x-12%29
f%281%29+=+sqrt%283%281%29-12%29
f%281%29+=+sqrt%283-12%29
f%281%29+=+sqrt%28-9%29 This result is NOT a real number since you have a negative under the square root. The answer is "does not exist" (assuming you haven't learned about complex/imaginary numbers yet).
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f%28x%29+=+sqrt%283x-12%29
f%28-1%29+=+sqrt%283%28-1%29-12%29
f%28-1%29+=+sqrt%28-3-12%29
f%28-1%29+=+sqrt%28-15%29 This result is NOT a real number since you have a negative under the square root. The answer is "does not exist" (assuming you haven't learned about complex/imaginary numbers yet).
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