Did I get these practice questions right?
1. Suppose that f is an even function whose domain is the set of all real numbers. Then which of the following can we claim to be true?
***The function f has an inverse f –1 that is even.
The function f has an inverse f –1, but we can't tell whether it's even or odd.
The function f has an inverse f –1, but we can't tell whether it's even or odd.
f –1 is not a function.
We can't tell whether or not f has an inverse.
2. Suppose that f is an odd function whose domain is the set of all real numbers. Then which of the following can we claim to be true?
(Points : 1)
The function f has an inverse f –1 that is even.
***The function f has an inverse f –1 that is odd.
The function f has an inverse f –1, but we can't tell whether it's even or odd.
The function f does not have an inverse.
We can't tell whether f has an inverse that's still a function.
#1 F^-1 is not a function
#2 we can't tell that bla bla
Hmm... Okay! I see how that works now! :) Thank you so much! I'll be posting a few more problems and would appreciate if you could help with those, too!
To determine if you got the practice questions right, I will walk you through the process of solving each question.
1. Suppose that f is an even function whose domain is the set of all real numbers. Then which of the following can we claim to be true?
To determine the answer, let's first review what it means for a function to be even. An even function is symmetric with respect to the y-axis, meaning it has the property f(x) = f(-x) for all x in the domain.
Now, we are asked to determine whether the function f has an inverse f^(-1) that is even. Remember that for a function to have an inverse, it must be both one-to-one (injective) and onto (surjective).
To check if f has an inverse, we need to ensure that it passes both the horizontal line test and the vertical line test. Since f is an even function, it is symmetric with respect to the y-axis. This means that no horizontal line crosses the graph of f at more than one point. Therefore, f is one-to-one.
Next, to determine if f is onto, we need to verify if every element in the range of f has a corresponding element in the domain. Since f has the domain of all real numbers, it means that every element in the range of f can be mapped to a unique real number. Therefore, f is onto.
Since f satisfies the conditions of being both one-to-one and onto, it has an inverse. However, whether the inverse will also be even or odd depends on the specific properties of the original function. The answer choices provided do not specify the properties of f, so we can't determine whether the inverse will be even or odd. Therefore, the correct answer is: "The function f has an inverse f^(-1), but we can't tell whether it's even or odd."
2. Suppose that f is an odd function whose domain is the set of all real numbers. Then which of the following can we claim to be true?
Similarly, to determine the answer, let's first review what it means for a function to be odd. An odd function is symmetric with respect to the origin, meaning it has the property f(x) = -f(-x) for all x in the domain.
Now, we are asked to determine whether the function f has an inverse f^(-1) that is odd. Remember that for a function to have an inverse, it must be both one-to-one (injective) and onto (surjective).
Since f is an odd function, it is symmetric with respect to the origin. This means that no vertical line crosses the graph of f at more than one point. Therefore, f is one-to-one.
Similar to question 1, we need to verify if every element in the range of f has a corresponding element in the domain to determine if f is onto. Since f has the domain of all real numbers, it means that every element in the range of f can be mapped to a unique real number. Therefore, f is onto.
Since f satisfies the conditions of being both one-to-one and onto, it has an inverse. Given that f is an odd function, we know that f^(-1) will also be odd. Therefore, the correct answer is: "The function f has an inverse f^(-1) that is odd."
Conclusion: Based on the explanations provided, you got both practice questions correct.
#1. f^-1 is not a function
Think of y = x^2
every value of y has two values of x.
#2 correct