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

[Rijad Hadzic]

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"

 

[Mark44]

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).

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.

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.

Homework Equations

The Attempt at a Solution

Plugged FoG into my calculator with x= -1. Got "undefined function"

 

[Rijad Hadzic]

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.

 

[Rijad Hadzic]

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 $\frac{1}{t(1+t)^{1/2}} - \frac 1 t$

later I get it down to

lim t -> 0 $\frac {t^2 -t - 1} {(t^2)(1+t)}$

lim t -> 0 $\frac {1 - \frac 1 t - \frac 1 t^2 } {1+t}$

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.

 

[Rijad Hadzic]

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..

 

[Mark44]

So say I have the problem (in Calculus)

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

later I get it down to

lim t -> 0 $\frac {t^2 -t - 1} {(t^2)(1+t)}$

lim t -> 0 $\frac {1 - \frac 1 t - \frac 1 t^2 } {1+t}$

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.

 

[Stephen Tashi]

lim t -> 0 $\frac {1 - \frac 1 t - \frac 1 t^2 } {1+t}$

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

How are you evaluating $\frac{1}{t}$ at $t = 0$ ?

 

[Mark44]

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.

 

[Rijad Hadzic]

Hi. I found the limit and it was actually -1/2

the original problem was

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

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.

 

[Mark44]

Hi. I found the limit and it was actually -1/2

Yes, that's what I got as well.