# Thread: Trouble solving a question: lim[x->-infty]f(x) * lim[x->+infty] < 0; find max of g(x)

1. ## Trouble solving a question: lim[x->-infty]f(x) * lim[x->+infty] < 0; find max of g(x)

Hello!
I'm new to the forum, so please excuse any mistakes I might be making.

I'm having trouble solving an exercise, I feel like there's something obvious I'm missing but can't see where...

There's a function which is continuous everywhere and $\lim_{x\to-\infty}f(x)\times\lim_{x\to+\infty}f(x)<0$ (both limits exist and are finite. I'm supposed to find the maximum of $g(x)=\dfrac{1}{1+\left[f(x)\right]^2}$

Since those limits have different signs, exist and are finite, I believe f has maximum and minimum (given by those two asymptotes)...

I got to the derivative of g easily, but I'm not sure what it gets me. $g'(x)=\dfrac{-2f(x)f'(x)}{\left(1+\left(f(x)\right)^2\right)}$ but since I don't know much about f, I can't say anything about the sign of g'... I've tried to find the second derivative, but in one attempt it was always positive, so it couldn't have a maximum...

Can anyone help me see what I'm missing? Thanks in advance!

2. Originally Posted by Eohyn
Hello!
I'm new to the forum, so please excuse any mistakes I might be making.

I'm having trouble solving an exercise, I feel like there's something obvious I'm missing but can't see where...

There's a function which is continuous everywhere and $\lim_{x\to-\infty}f(x)\times\lim_{x\to+\infty}f(x)<0$ (both limits exist and are finite. I'm supposed to find the maximum of $g(x)=\dfrac{1}{1+\left[f(x)\right]^2}$

Since those limits have different signs, exist and are finite, I believe f has maximum and minimum (given by those two asymptotes)...

I got to the derivative of g easily, but I'm not sure what it gets me. $g'(x)=\dfrac{-2f(x)f'(x)}{\left(1+\left(f(x)\right)^2\right)}$ but since I don't know much about f, I can't say anything about the sign of g'... I've tried to find the second derivative, but in one attempt it was always positive, so it couldn't have a maximum...

Can anyone help me see what I'm missing? Thanks in advance!
You are on the wrong track altogether. You cannot use derivatives because you do not know that f(x) is everywhere differentiable. (Assuming it is differentiable, you would still need to ensure that g(x) is everywhere differentiable and calculate the derivative correctly. Using the power rule might suggest squaring the denominator of g(x) when taking the first derivative.) Moreover, you cannot assume that the asymptotes are reached as x increases or decreases without bound. Nor can you assume that f(x) does not have absolute values greater than the absolute value of the asymptotes at some finite values of x.

What you do know is that the asymptotes have opposite sign so f(x) has at least one zero. (Intermediate Value Theorem)

$f(a) = 0 \implies \{f(a)\}^2 = 0 \implies g(a) = 1.$

$f(b) \ne 0 \implies \{f(b)\}^2 > 0 \implies g(b) < 1.$

Now what do you think the maximum of g(x) is?

Differential calculus is a very powerful tool, but not all problems can take advantage of it.

3. Originally Posted by JeffM
You are on the wrong track altogether. You cannot use derivatives because you do not know that f(x) is everywhere differentiable. (Assuming it is differentiable, you would still need to ensure that g(x) is everywhere differentiable and calculate the derivative correctly. Using the power rule might suggest squaring the denominator of g(x) when taking the first derivative.) Moreover, you cannot assume that the asymptotes are reached as x increases or decreases without bound. Nor can you assume that f(x) does not have absolute values greater than the absolute value of the asymptotes at some finite values of x.

What you do know is that the asymptotes have opposite sign so f(x) has at least one zero. (Intermediate Value Theorem)

$f(a) = 0 \implies \{f(a)\}^2 = 0 \implies g(a) = 1.$

$f(b) \ne 0 \implies \{f(b)\}^2 > 0 \implies g(b) < 1.$

Now what do you think the maximum of g(x) is?

Differential calculus is a very powerful tool, but not all problems can take advantage of it.
First of all, thanks for the help!

You're right, I can't assume the function is differentiable or that the asymptotes are the maximum and minimum... Silly me!

If I'm understanding correctly, then g has a maximum when $x:f(x)=0$. I know such x exists because of the Intermediate Value Theorem. Another question: can I guarantee there's only one such x?

4. Originally Posted by Eohyn
First of all, thanks for the help!

You're right, I can't assume the function is differentiable or that the asymptotes are the maximum and minimum... Silly me!

If I'm understanding correctly, then g has a maximum when $x:f(x)=0$. I know such x exists because of the Intermediate Value Theorem. Another question: can I guarantee there's only one such x?
You cannot guarantee that f(x) has a unique zero. But you do not need to do so. You are asked to find the maximum of g(x). Its value will be 1 at EVERY x such that f(x) = 0. The value of g(x) will be less than 1 but more than zero at every x such that f(x) $\ne$ 0. Consequently the maximum value of g(x) is 1 no matter how many zeroes f(x) may have.

5. Originally Posted by JeffM
You cannot guarantee that f(x) has a unique zero. But you do not need to do so. You are asked to find the maximum of g(x). Its value will be 1 at EVERY x such that f(x) = 0. The value of g(x) will be less than 1 but more than zero at every x such that f(x) $\ne$ 0. Consequently the maximum value of g(x) is 1 no matter how many zeroes f(x) may have.
Thank you, you've been a great help!

6. Hello,

Just one final question. Do you think it's necessary to prove that I can find a closed and limited interval from a to b and formally prove such an interval is bounded? Personally, I don't think it's needed since the function is continuous everywhere...

7. Originally Posted by Eohyn
Hello,

Just one final question. Do you think it's necessary to prove that I can find a closed and limited interval from a to b and formally prove such an interval is bounded? Personally, I don't think it's needed since the function is continuous everywhere...

8. Originally Posted by Eohyn
I don't understand the question. The interval [a, b] is bounded; that is what the notation means.

9. Originally Posted by JeffM
I don't understand the question. The interval [a, b] is bounded; that is what the notation means.
That's what I thought, but the teacher is saying unless you prove the interval is closed and bounded, the answer is very incomplete...

10. Originally Posted by Eohyn
That's what I thought, but the teacher is saying unless you prove the interval is closed and bounded, the answer is very incomplete...
It will help if you show us the exact answer you submitted, so we can see what might have been omitted. As it is, it is not clear what a and b you are referring to.

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