When f(x) =0

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Loki123

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This might be stupid to ask, but I really want to check If I did this correctly. I am concerned that the absolute somehow influences the solution. IMG_20220503_165130.jpg
 
You have the condition that [imath]x \neq 1[/imath]. Why are you saying that the domain is x > -1?

Again, can x = -20?

-Dan
 
1651591326117.png

Think very carefully! What are you saying is false here? Are you saying that no value is excluded from the domain, or that no value is in the domain? Or something else?

You probably mean the right thing, though it is not entirely clear, because you omitted details of your reasoning. If so, what is your final answer to the problem?

Incidentally, another way to approach this would be just to check each of your originally claimed zeros by putting them into the function.
 
You are, I think, allowing yourself to get tied up unnecessarily by your uncertainty about absolute values. I do not understand EXACTLY what your writing is supposed to mean.

The basic idea is that a function with absolute values is effectively defined piecewise.

[math]\text {sgn}(x + 1) = - 1 \implies |x + 1| = - (x + 1) \implies\\ f(x) = \dfrac{x(x + 1)(x + 2)(x + 3)}{x + 1 + |x + 1|} = \dfrac{x(x + 1)(x + 2)(x + 3)}{x + 1 - (x + 1)} \not \in \mathbb R.[/math]
So we can say that this function does not exist where

[math]x + 1 < 0 \implies x < - 1.[/math]
I think that is what you were getting at.

[math]\text {sgn}(x + 1) = 1 \implies |x + 1| = (x + 1) \implies\\ f(x) = \dfrac{x(x + 1)(x + 2)(x + 3)}{x + 1 + |x + 1|} = \dfrac{x(x + 1)(x + 2)(x + 3)}{x + 1 + (x + 1)} =\\ \dfrac{x(x + 1)(x + 2)(x + 3)}{2(x + 1)}[/math].

You have no domain issues if [imath]x + 1 > 0 \implies x > - 1[/imath].

What happens if x = - 1?
 
topsquark said:
You have the condition that [imath]x \neq 1[/imath]. Why are you saying that the domain is x > -1?

Again, can x = -20?

-Dan
Click to expand...
x can't be -20 because if x<-1 we are diving by 0.
the domain has to be x>=-1, but we exclude -1 because it's 0 for that too. So x>-1
 
Dr.Peterson said:
View attachment 32540

Think very carefully! What are you saying is false here? Are you saying that no value is excluded from the domain, or that no value is in the domain? Or something else?

You probably mean the right thing, though it is not entirely clear, because you omitted details of your reasoning. If so, what is your final answer to the problem?

Incidentally, another way to approach this would be just to check each of your originally claimed zeros by putting them into the function.
Click to expand...
if we have x<-1, we are diving by 0 which is not possible.
 
JeffM said:
You are, I think, allowing yourself to get tied up unnecessarily by your uncertainty about absolute values. I do not understand EXACTLY what your writing is supposed to mean.

The basic idea is that a function with absolute values is effectively defined piecewise.

[math]\text {sgn}(x + 1) = - 1 \implies |x + 1| = - (x + 1) \implies\\ f(x) = \dfrac{x(x + 1)(x + 2)(x + 3)}{x + 1 + |x + 1|} = \dfrac{x(x + 1)(x + 2)(x + 3)}{x + 1 - (x + 1)} \not \in \mathbb R.[/math]
So we can say that this function does not exist where

[math]x + 1 < 0 \implies x < - 1.[/math]
I think that is what you were getting at.

[math]\text {sgn}(x + 1) = 1 \implies |x + 1| = (x + 1) \implies\\ f(x) = \dfrac{x(x + 1)(x + 2)(x + 3)}{x + 1 + |x + 1|} = \dfrac{x(x + 1)(x + 2)(x + 3)}{x + 1 + (x + 1)} =\\ \dfrac{x(x + 1)(x + 2)(x + 3)}{2(x + 1)}[/math].

You have no domain issues if [imath]x + 1 > 0 \implies x > - 1[/imath].

What happens if x = - 1?
Click to expand...
if it's x=-1 we are dividing by 0. So the domain is x>-1
 
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