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C3 Functions - Am I right? Need a Maths whizz here! watch

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    The question asks:
    Determine\ t^-1(x) if\ the\ function\ t(x)\ is\ described\ by\ t(x) = x^2 - 6x + 5 [x \in \mathbb{R} , x \geq 5

    This is what I've done so far:
    Let y = x^2 - 6x + 5
     y = (x - 3)^2 - 4
     x = \sqrt{y + 4} + 3
     t^-1(x) = \sqrt{x + 4} + 3 \ and\ [x \in \mathbb{R} , x \geq -4

    I got the domain as -4, which makes sense, but my book says it is 0. Is this a typo, am I wrong...help please? :confused:

    Oh and also, would the actual domain of  t(x) be any real value (since you can have any value of x) and would the range be for  x \geq -4 ?

    This would mean that the domain of t^-1(x) would be -4 and the range would be greater than 3...but then that would mean the domain of t(x) would have to be 3, due to that rule that the domain of f(x) = range of f^-1(x). If you were looking at the function t(x) as  t(x) = (x - 3)^2 - 4 , then it would make sense that the doman has to be 3, but then you could also look at the function as  t(x) = x^2 - 6x + 5 .

    I know I'm waffling now, but I seriously do not understand! God, I am so confused! Can someone please put my out of misery and I will + rep!
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    (Original post by Narik)
    The question asks:
    Determine\ t^-1(x) if\ the\ function\ t(x)\ is\ described\ by\ t(x) = x^2 - 6x + 5 [x \in \mathbb{R} , x \geq 5

    This is what I've done so far:
    Let y = x^2 - 6x + 5
     y = (x - 3)^2 - 4
     x = \sqrt{y + 4} + 3
     t^-1(x) = \sqrt{x + 4} + 3 \ and\ [x \in \mathbb{R} , x \geq -4

    I got the domain as -4, which makes sense, but my book says it is 0. Is this a typo, am I wrong...help please? :confused:

    Oh and also, would the actual domain of  t(x) be any real value (since you can have any value of x) and would the range be for  x \geq -4 ?

    This would mean that the domain of t^-1(x) would be -4 and the range would be greater than 3...but then that would mean the domain of t(x) would have to be 3, due to that rule that the domain of f(x) = range of f^-1(x). If you were looking at the function t(x) as  t(x) = (x - 3)^2 - 4 , then it would make sense that the doman has to be 3, but then you could also look at the function as  t(x) = x^2 - 6x + 5 .

    I know I'm waffling now, but I seriously do not understand! God, I am so confused! Can someone please put my out of misery and I will + rep!
    Ok I don't know enough about domain and range to fully explain this to you, but I do know that the book is correct in this case.

    Note: Read all my > as greater than or equal to signs, cba to put in latex

    If you put x = 5 into your expression for t(x), you'll find this gives 0. Since 5 is the lowest allowed value of x, and the function keeps increasing for all x>5, this means that x = 5 will map to the lowest possible value for t(x). Therefore the range of t(x) in this case will be t(x) > 0. Using the rule that the range of the function is the domain of the inverse, this means that the domain of the inverse function is actually 0 as your book says.

    The domain of t(x) has been specified in the question as x>5, therefore the range of the inverse is x>5 as the range of an inverse is the domain of the function.

    Were you to be allowed any x value, the domain for t(x) would indeed be all x as you say. In this case the range would be found by finding the minimum of the quadratic (via differentiation) or by sketching it. It would be x > -4 like you say.

    I think that actually you do understand the topic, just didnt take into account the restricted domain of x>5!
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    (Original post by Narik)
    The question asks:
    Determine\ t^-1(x) if\ the\ function\ t(x)\ is\ described\ by\ t(x) = x^2 - 6x + 5 [x \in \mathbb{R} , x \geq 5

    This is what I've done so far:
    Let y = x^2 - 6x + 5
     y = (x - 3)^2 - 4
     x = \sqrt{y + 4} + 3
     t^-1(x) = \sqrt{x + 4} + 3 \ and\ [x \in \mathbb{R} , x \geq -4

    I got the domain as -4, which makes sense, but my book says it is 0. Is this a typo, am I wrong...help please? :confused:

    Oh and also, would the actual domain of  t(x) be any real value (since you can have any value of x) and would the range be for  x \geq -4 ?

    This would mean that the domain of t^-1(x) would be -4 and the range would be greater than 3...but then that would mean the domain of t(x) would have to be 3, due to that rule that the domain of f(x) = range of f^-1(x). If you were looking at the function t(x) as  t(x) = (x - 3)^2 - 4 , then it would make sense that the doman has to be 3, but then you could also look at the function as  t(x) = x^2 - 6x + 5 .

    I know I'm waffling now, but I seriously do not understand! God, I am so confused! Can someone please put my out of misery and I will + rep!

    as you are given that x > 5 (domain), you know that the inverse function must also be greater then 5.

    hence x>0

    this is as the inverse codomain is equal to that of the domain of the original function. also the domain of the inverse function is the codomain of the function. so you could check that the codomain of the function is x>0 (which it is!)
 
 
 
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