Does x = -1 Make the Function FoG Undefined Even If Part of It Equals Zero?

In summary, a rational function is a fraction where the numerator and denominator are polynomial expressions. To add two rational functions, a common denominator is found and the numerators are added together, then simplified if possible. Any two rational functions can be added, but the resulting function may not always be in rational form. The purpose of adding two rational functions is to combine them into a single function for solving problems or simplifying expressions. There are two special cases when adding two rational functions: when the denominators are the same, and when the functions are reciprocals. These cases allow for direct addition or subtraction of the numerators.
  • #1
Rijad Hadzic
321
20

Homework Statement


Say you have a fn

FoG = x+2/x+1 + x+1/x+2

if x = -1, the first one is undefined. But the second one would end up as 0, a real number

I'm trying to understand, would x = -1 make the whole term FoG undefined? because only x+2/x+1 would be undefined, x+1/x+2 would be = to 0, so why isn't the answer zero, instead of undefined?Also one more question so I don't have to make threads.

I'm freshening up on my calculus.

" Express the function in the for FoGoH: ( sec(x^(1/2)) ) ^4 "

the book did this:
h(x) = x^(1/2) g (x) = sec(x) f(x) = x^4

but I did: h(x) = x g(x) = x^1/2 f(x) = (sec(x) ) ^4

you get the same answer... surely I'm not wrong am I?

Homework Equations

The Attempt at a Solution


Plugged FoG into my calculator with x= -1. Got "undefined function"
 
Last edited:
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  • #2
Rijad Hadzic said:

Homework Statement


Say you have a fn

FoG = x+2/x+1 + x+1/x+2
Please use parentheses. I know what you meant, but what you wrote on the right side is ##x + \frac 2 x + 1 + x + \frac 1 x + 2##
At the very least the right side should be written as (x + 2)/(x + 1) + (x + 1)/(x + 2).
Rijad Hadzic said:
if x = -1, the first one is undefined. But the second one would end up as 0, a real number
But the sum is undefined. For a sum to be defined (i.e., equal to a number), both terms in the sum have to be defines.
Rijad Hadzic said:
I'm trying to understand, would x = -1 make the whole term FoG undefined? because only x+2/x+1 would be undefined, x+1/x+2 would be = to 0, so why isn't the answer zero, instead of undefined?Also one more question so I don't have to make threads.

I'm freshening up on my calculus.

" Express the function in the for FoGoH: ( sec(x^(1/2)) ) ^4 "

the book did this:
h(x) = x^(1/2) g (x) = sec(x) f(x) = x^4

but I did: h(x) = x g(x) = x^1/2 f(x) = (sec(x) ) ^4

you get the same answer... surely I'm not wrong am I?
Your f(x) is itself a composite function, the same as their f(g(x)).
It's sort of a waste of effort to write h(x) = x as you did.
Rijad Hadzic said:

Homework Equations

The Attempt at a Solution


Plugged FoG into my calculator with x= -1. Got "undefined function"
 
  • #3
Appreciate the reply. Sorry for the laziness, I'll write a better OP next time.

Your reply to all my points makes complete sense. Thank you.
 
  • #4
Mark44 said:
But the sum is undefined. For a sum to be defined (i.e., equal to a number), both terms in the sum have to be defines.

So say I have the problem (in Calculus)

lim t->0 [itex] \frac{1}{t(1+t)^{1/2}} - \frac 1 t [/itex]

later I get it down to

lim t -> 0 [itex] \frac {t^2 -t - 1} {(t^2)(1+t)} [/itex]

lim t -> 0 [itex] \frac {1 - \frac 1 t - \frac 1 t^2 } {1+t} [/itex]

and then evaluate for t = 0, why would my answer of 1/1 be valid if the sum is not defined?

Sorry for bumping this old thread. Not sure if making a new one would have been appropriate.
 
  • #5
Damn on checking that answer it looks like my answer evaluated isn't 1/1... my question in regards to evaluating a 1/x function where x -> 0 still holds but Ill be back with the right method for getting the answer..
 
  • #6
Rijad Hadzic said:
So say I have the problem (in Calculus)

lim t->0 [itex] \frac{1}{t(1+t)^{1/2}} - \frac 1 t [/itex]

later I get it down to

lim t -> 0 [itex] \frac {t^2 -t - 1} {(t^2)(1+t)} [/itex]

lim t -> 0 [itex] \frac {1 - \frac 1 t - \frac 1 t^2 } {1+t} [/itex]

and then evaluate for t = 0, why would my answer of 1/1 be valid if the sum is not defined?
No, not at all. Two of the terms in the numerator are not defined, which makes the whole numerator undefined.
I think you have made a mistake going from the original limit to the next line. I get a negative value for this limit.
 
  • #7
Rijad Hadzic said:
lim t -> 0 [itex] \frac {1 - \frac 1 t - \frac 1 t^2 } {1+t} [/itex]

and then evaluate for t = 0, why would my answer of 1/1

How are you evaluating ##\frac{1}{t}## at ##t = 0## ?
 
  • #8
Looking at the limit in post #5, you have ##\lim_{t \to 0}\frac 1 {t \sqrt{1 + t}} - \frac 1 t##
One of the properties of limits is that ##\lim_{x \to a} f(x) + g(x) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)##, provided that both limits on the right exist. In the limit above, ##\lim_{t \to 0} \frac 1 t## doesn't exist, so it's not valid to split the expression you started with into two separate limits.

It is possible, though, in this example to do some algebraic manipulation to get something whose limit does exist.
 
  • #9
Hi. I found the limit and it was actually -1/2

the original problem was

lim t -> 0 [itex] \frac {1}{t(1+t)^(1/2)} - \frac 1 t [/itex]

It was just a a long process of using the conjugate formula.

I should have just taken it for a fact that if you get 1/x evaluated with x = 0 that it is going to be undefined and that is your answer.

Solving my problem it didn't even require me to have anything undefined.

Of course this makes sense because the properties of algebra aren't just magically going to change. I was a fool for trying to rush the problem.

Thank you guys for the consistent help.
 
  • #10
Rijad Hadzic said:
Hi. I found the limit and it was actually -1/2
Yes, that's what I got as well.
 

FAQ: Does x = -1 Make the Function FoG Undefined Even If Part of It Equals Zero?

1. What is a rational function?

A rational function is a mathematical expression that can be written as the ratio of two polynomial functions. In other words, it is a fraction where the numerator and denominator are polynomial expressions.

2. How is addition of 2 rational functions performed?

To add 2 rational functions, we need to first find a common denominator, then add the numerators together. Finally, we simplify the resulting fraction, if possible.

3. Can any rational function be added to another rational function?

Yes, as long as both functions are in the form of a ratio of polynomial expressions, they can be added together. It is important to note that the resulting function may not always be in the form of a rational function.

4. What is the purpose of adding 2 rational functions?

The purpose of adding 2 rational functions is to combine them into a single rational function that represents the sum of the two original functions. This can be useful in solving mathematical problems or simplifying complex expressions.

5. Are there any special cases when adding 2 rational functions?

Yes, there are two special cases when adding 2 rational functions: when the denominators are the same, and when the functions are reciprocal of each other. In these cases, the addition can be simplified by adding or subtracting the numerators directly.

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