Tell which of the following statements are equivalent to f(x) = 0

[acemi123]

I didn't understand what does this question means. Does f(x) = 0 means, put 0 instead of x and find solution?
Please, can someone explain to me how to solve?

Let f be a real map of one real variable. Tell which of the following statements
are equivalent to f(x) = 0:

1. f(x) + x = x
2. f(x) + 1 / (x^2−3) = 1
3. (x^2 − 1).f(x) = 0
4. (x^2 + 1).f(x) = 0
5. f^2(x) = 0
6. f(x) − x^2 = x^4
7. f(x) / (x^4+1) = 0
8. f(x) / (x−3) = 0

 

[Otis]

… Does f(x) = 0 [mean to] put 0 instead of x …

No. If they wanted zero to be the input, then they would've written f(0).

It seems like you've forgotten the meaning of function notation.

Symbol f(x) is a variable. It's just a different way of writing y (called 'function notation').

Does it help, if you write y instead of f(x)?

​

Let y be a real map of one real variable. Tell which of the following statements​

are equivalent to y = 0:​

​

1. y + x = x​

2. y + 1 / (x^2−3) = 1​

3. (x^2 − 1)*y = 0​

4. (x^2 + 1)*y = 0​

5. y^2 = 0​

6. y − x^2 = x^4​

7. y / (x^4+1) = 0​

8. y / (x−3) = 0​

​

?

 

[acemi123]

Thanks for the reply. But the answer is given as 1,4,5, and 7.
However, according to what you say when I left y alone some answers are not in correct answers. For example number 3 and 8 why not correct?
Thanks.

 

[JeffM]

But Otis did not tell you [MATH]ANYTHING[/MATH] like that.. You added that for some reason.

We start by saying [MATH]y = f(x).[/MATH]
Let's look at # 1.

[MATH]f(x) + x = x \implies y + x = x \implies y = 0 \implies f(x) =0.[/MATH]
So in the case of #1, f(x) = 0 for any real value of x is implied. That agrees with your book.

Let's look at # 3.

[MATH](x^2 - 1) * f(x) = 0 \implies (x^2 - 1) * y = 0 \implies[/MATH]
[MATH]y = 0 \text { or } x = 1 \text { or } x = -\ 1.[/MATH]
Now it is possible that f(x) = 0 for every value of x in this case. But it is NOT NECESSARILY true.

[MATH]f(-\ 1) = 50 \implies (x^2 - 1) * f(x) = \{(-\ 1)^2 - 1\} * 50 = (1 - 1) * 50 = 0 * 50 = 0.[/MATH]
[MATH]f(1) = 500 \implies (x^2 - 1) * f(x) = (1^2 - 1) * 50 = (1 - 1) * 50 = 0 * 50 = 0.[/MATH]
So you cannot conclude that

[MATH]f(x) = 0 \text { for all real values of x} \iff (x^2 - 1) * f(x) = 0.[/MATH]
I'll bet dollars to doughnuts that the book posed the question somewhat like "equivalent to f(x) = 0 for all real values of x." This is the reason why it is so important to give your problem completely and exactly.

So again the book is correct.

 

[Deleted member 4993]

"dollars to doughnuts " - days are coming when 1 dollar will not be enough for 1 doughnut ........

 

[acemi123]

But Otis did not tell you [MATH]ANYTHING[/MATH] like that.. You added that for some reason.

We start by saying [MATH]y = f(x).[/MATH]
Let's look at # 1.

[MATH]f(x) + x = x \implies y + x = x \implies y = 0 \implies f(x) =0.[/MATH]
So in the case of #1, f(x) = 0 for any real value of x is implied. That agrees with your book.

Let's look at # 3.

[MATH](x^2 - 1) * f(x) = 0 \implies (x^2 - 1) * y = 0 \implies[/MATH]
[MATH]y = 0 \text { or } x = 1 \text { or } x = -\ 1.[/MATH]
Now it is possible that f(x) = 0 for every value of x in this case. But it is NOT NECESSARILY true.

[MATH]f(-\ 1) = 50 \implies (x^2 - 1) * f(x) = \{(-\ 1)^2 - 1\} * 50 = (1 - 1) * 50 = 0 * 50 = 0.[/MATH]
[MATH]f(1) = 500 \implies (x^2 - 1) * f(x) = (1^2 - 1) * 50 = (1 - 1) * 50 = 0 * 50 = 0.[/MATH]
So you cannot conclude that

[MATH]f(x) = 0 \text { for all real values of x} \iff (x^2 - 1) * f(x) = 0.[/MATH]
I'll bet dollars to doughnuts that the book posed the question somewhat like "equivalent to f(x) = 0 for all real values of x." This is the reason why it is so important to give your problem completely and exactly.

So again the book is correct.

Thanks, Jeff, but what about number 4 ( (x^2 + 1)*y = 0 ) . It is the same situation just the sign is different. what has changed? why this one is correct but number 3 is not?
Thanks lot.

 

[Deleted member 4993]

Thanks, Jeff, but what about number 4 ( (x^2 + 1)*y = 0 ) . It is the same situation just the sign is different. what has changed? why this one is correct but number 3 is not?
Thanks lot.

Can you think of a value for x for which x² + 1 = 0 ?