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thereddevils
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Homework Statement
what is the condition for function fg^(-1) to exist ?
Homework Equations
The Attempt at a Solution
i understand that g^(-1) only happens when its one-one , how about fg^(-1) ?
thereddevils said:Homework Statement
what is the condition for function fg^(-1) to exist ?
Homework Equations
The Attempt at a Solution
i understand that g^(-1) only happens when its one-one , how about fg^(-1) ?
HallsofIvy said:[itex]fg^{-1}[/itex] is only defined on the intersection of the domains of f and g.
Susanne217 said:We know that HallsoftIvy, but you think since this is a pre-Calculus question that the original poster is simply looking for and try to understand the definition of
[tex]f \circ g^{-1}[/tex]?
You aren't being clear in what you're asking. Which of the following are you asking about?thereddevils said:Homework Statement
what is the condition for function fg^(-1) to exist ?
Homework Equations
The Attempt at a Solution
i understand that g^(-1) only happens when its one-one , how about fg^(-1) ?
Mark44 said:You aren't being clear in what you're asking. Which of the following are you asking about?
- fg-1, the product of f and g-1
- (fg)-1, the inverse of the product of f and g
- f [itex]\circ[/itex] g-1, the composition of f and g-1
- (f [itex]\circ[/itex] g)-1, the inverse of the composition of f and g
thereddevils said:i understand that g^(-1) only happens when its one-one
Mark44 said:When who is one-to-one? See if you can restate your question in a more precise way.
Mark44 said:For this example, the restricted domain for g makes it a one-to-one function, so the inverse, g-1, exists.
For g-1 o f, you need to look at the domain of f (x such that x > 0), and the range of f. The outputs from f are going to be the inputs to g-1. Are there any numbers in the range of the ln function that are not allowed in the domain for g-1?
Mark44 said:Yes. Now, can you figure out how to restrict the domain of f so that the range of f is {y | y > -1}?
IOW, what restrictions can you place on the domain of ln(x) so that the range is {y | y > -1}?
A function is a mathematical relationship between two quantities, where each input (or independent variable) has only one corresponding output (or dependent variable).
In this function, f and g represent two separate functions. The f function is applied first to the input, and then the g function is applied to the result of the f function. The inverse function of g is then applied to the final result.
The condition for fg^(-1) to exist is that both f and g must be one-to-one functions. This means that each input has only one corresponding output, and no two inputs can have the same output. This ensures that the inverse function of g exists.
Having the inverse function allows us to "undo" the original function and retrieve the original input. This is useful in many mathematical applications, such as solving equations, finding the domain and range of a function, and creating inverse trigonometric functions.
If the condition is not met, then fg^(-1) does not exist and the function cannot be evaluated. This could be due to the functions f and g not being one-to-one, or the inverse function of g not existing. In this case, the function may be undefined or may have multiple outputs for a single input.