SOLUTION: I am still unsure when it comes to functions. I have a question that is as follows: Is f(x) = x^2 + 2x - 3 a one-to-one function? Explain your answer. When figuring out the p

Algebra ->  Functions -> SOLUTION: I am still unsure when it comes to functions. I have a question that is as follows: Is f(x) = x^2 + 2x - 3 a one-to-one function? Explain your answer. When figuring out the p      Log On


   

Question 213488: I am still unsure when it comes to functions. I have a question that is as follows: Is f(x) = x^2 + 2x - 3 a one-to-one function? Explain your answer.
When figuring out the problem if f(a) = f(b) then my answer would be no as
a^2 + 2a - 3 = b^2 + 2b -3 does not factor down to a = b.
Am I figuring this correctly?
Thank you = Lori
Found 2 solutions by Theo, jim_thompson5910:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
if every value of x yield one and only one value of y, the equation is a function.
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f(x) = x^2 + 2x - 3 is a function.
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let y = f(x) and we can graph this equation.
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graph+%28300%2C300%2C-10%2C10%2C-10%2C10%2Cx%5E2+%2B+2x+-+3%29
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if f(a) equals f(b), this does not mean the equation is not a function.
you would have one value of y for 2 values of x.
f(x1) would be y
f(x2) would be y
you still only have one value of y for each value of x so the equation is still a function.
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the definition of a function is one value of f(x) for each value of x.
f(a) = f(b) does not violate this.
take y = x^2
let a = 2
then f(a) = 4
let b = -2
then f(b) = 4
f(2) = f(-2)
you have f(a) = f(b) but you still only have one value of y for each value of x.
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Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
You're on the right track. A function is one-to-one if f%28a%29=f%28b%29 only means a=b. In other words, if two outputs of a function are equal, then the inputs must be equal for the function to be one-to-one.


So if you think that this function is NOT one-to-one, then simply pick a counter-example to prove this claim false. Let's say that f%28a%29=f%28b%29=-3. Plugging this into f%28x%29=x%5E2%2B2x-3 gets us -3=x%5E2%2B2x-3. Now solve for 'x' to get x=0 or x=-2


So if we plug in either 0 or -2 into f(x), we'll get -3 as an output. Ie. f%280%29=f%28-2%29=-3. Since 0%3C%3E-2, this means that the function is NOT one-to-one. So you are correct.