Ratio of functions, surjective (analysis course)

• riskandar
In summary: The intermediate value theorem states that if a continuous function takes on values of both very large and very large negative, then it must take on every value in between. So to prove g is surjective, you just need to show that it takes on both very large and very large negative values, which is what the problem asks for.In summary, the conversation discusses a proof involving a continuous function f and a function g defined as g(x) = x^k + f(x), where k is an integer greater than or equal to 1. The problem asks to prove that g is surjective if k is odd and that there exists a real number y such that the image of g is [y,inf) if k is even. The
riskandar

Homework Statement

let f: R->R be a continuous function
Suppose k>=1 is an integer such that

lim f(x)/x^k = lim f(x)/x^k = 0
x->inf x->-inf

set g(x)= x^k + f(x)

g: R->R

Prove that
(i) if k is odd, then g is surjective
(ii) if k is even, then there is a real number y such that the image of g is [y,inf)

The Attempt at a Solution

I am completely stuck at this all I can think of is x^k goes to infinity then the ratio of the functions can go to 0 if either f(x) goes to 0 or f(x) is a constant or f(x) goes to infinity slower than x^k (I am not sure about this)

Any help will be very much appreciateve

Because $$g$$ is continuous, you can use the intermediate value theorem. To show that $$g$$ is surjective, it is enough to show that $$g$$ becomes both very large and very large negative.

To prove this part, you need to think about the qualitative behavior of $$f$$ and $$g$$. The hypothesis on $$f$$ says that $$f(x)$$ is negligible compared to $$x^k$$ as $$|x|$$ becomes large. Therefore $$g(x)$$ should "behave almost like" $$x^k$$ as $$|x|$$ becomes large. Figure out a way to make this precise.

ystael said:
Because $$g$$ is continuous, you can use the intermediate value theorem. To show that $$g$$ is surjective, it is enough to show that $$g$$ becomes both very large and very large negative.

To prove this part, you need to think about the qualitative behavior of $$f$$ and $$g$$. The hypothesis on $$f$$ says that $$f(x)$$ is negligible compared to $$x^k$$ as $$|x|$$ becomes large. Therefore $$g(x)$$ should "behave almost like" $$x^k$$ as $$|x|$$ becomes large. Figure out a way to make this precise.

Where do I use the intermediate value theorem? Is it to prove surjective?

riskandar said:
Where do I use the intermediate value theorem? Is it to prove surjective?

Yes.

1. What does it mean for a function to be surjective?

A function is surjective if every element in the range (output) of the function has at least one corresponding input value in the domain. In other words, the function "maps onto" its entire range.

2. How is the ratio of two functions calculated?

The ratio of two functions is calculated by dividing the output of the first function by the output of the second function for a given input value. This can also be represented algebraically as f(x)/g(x).

3. Can the ratio of two surjective functions ever be undefined?

Yes, the ratio of two surjective functions can be undefined if the denominator function has an output value of 0 for a given input value. This would result in a division by 0 error, making the ratio undefined.

4. How does the surjectivity of a function affect its ratio with another function?

If both functions are surjective, then their ratio will also be surjective. However, if only one of the functions is surjective, the ratio may not be surjective as there may be elements in the range of the numerator function that do not have a corresponding input value in the range of the denominator function.

5. Can a function be both surjective and non-surjective at the same time?

No, a function cannot be both surjective and non-surjective at the same time. A function is either surjective or it is not. If a function is surjective, it means that it maps onto its entire range, so it cannot have elements in its range that do not have a corresponding input value in its domain.

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