# Calculate (and argue) the critical points of an exponential function

• mcastillo356
In summary, the conversation discusses the function ##x^x## and its properties. It is proven that the domain of ##x^x## is ##(0, \infty)## and the question of why ##0^0## is not defined as 0 is raised. The conversation also explores the possibility of defining ##0^0## as 1, but the properties of such a definition are not specified. The conversation ends with a note of gratitude for the help and discussion on the topic.
mcastillo356
Gold Member
Homework Statement
Calculate the critical point of ##y=x^x##
Relevant Equations
Chain Rule, Product Rule
Hi everybody

We can't differentiate ##x^x## neither like a power function nor an exponential function. But ##x^x=e^{x\mbox{ln}x}##. So

##\dfrac{d}{dx}x^x=\dfrac{d}{dx}e^{x\mbox{ln}x}=x^x(\mbox{ln}x+1)##

And here comes the doubt: prove the domain of ##x^x## is ##(0, +\infty)##

Why is only positive?; why it will never be 0?

So ##(\mbox{ln}x+1)=0\Rightarrow{x=1/e}##

It's a solved exercise at Calculus, by R.A.A. It's the only question not solved (the domain, I mean).

You can define ##0^0## if you want, but fractional powers of negative numbers are undefined, or require complex numbers. E.g. ##-0.5^{-0.5}## is not a real number.

mcastillo356
Can ##0^0=1##? The book says: ##x^x## is never 0. This is the graph:

##0^x=0## on one hand; ##x^0=1##. This is all I can provide.

##0^0## is given a value per definition. If you want continuity of ##x \longrightarrow x^0## at ##x=0## then you have to set ##0^0=1,## and if you want continuity of ##x\longrightarrow 0^x## at ##x=0## then you have to set ##0^0=0.##

My personal opinion is, that only ##0^0=1## is a valid definition because exponentiation is an abbreviation for multiplication and all multiplications over an empty set of indexes equals the neutral element of multiplication, which is ##1.##
$$\sum_{i\in \{\}} a_i = 0 \quad ;\quad \prod_{i\in \{\}} a_i = 1$$

Last edited:
mcastillo356
The next step would be to prove ##\displaystyle\lim_{x \to{0+}}{x^x}=f(0)=1##? I am just wondering.

mcastillo356 said:
The next step would be to prove ##\displaystyle\lim_{x \to{0+}}{x^x}=f(0)=1##? I am just wondering.
Yes. E.g. we get for ##x>0##
$$\lim_{n \to \infty}\left(\dfrac{1}{n}\right)^{1/n}= \lim_{n \to \infty}\left(1-\dfrac{\log n}{n}+O(n^{-2})\right)=1.$$
But if you decouple base and exponent, say ##f(x,y)=x^y,## then all depends on which path you approach ##(0,0).## That is where the question: "Why isn't ##0^0=0## ?" comes from. ##x\longmapsto x^x ## and ##(x,y)\longmapsto x^y## are different functions with different properties. And even ##f(x,y)## depends on its domain: real positive numbers, all real numbers and complex values, complex numbers and complex values. These are questions that should be answered before we discuss ##0^0.## The answer depends on them.

mcastillo356
##\displaystyle\lim_{x \to{0^+}}{e^{x\ln(x)}}=e^{\lim_{x \to{0^+}}{x\ln(x)}}=e^{\lim_{x \to{0^+}}{\frac{\ln(x)}{1/x}}}##, and, by L'Hopital, ##\displaystyle e^{\lim_{x \to{0^+}}{-x}}=1##.

Is this equal to ##f(0)##? Don't know. ##x^x=e^{x\ln(x)}##. Does this alow me to say "no", because ##\ln 0## is nonsense?.

mcastillo356 said:
##\displaystyle\lim_{x \to{0^+}}{e^{x\ln(x)}}=e^{\lim_{x \to{0^+}}{x\ln(x)}}=e^{\lim_{x \to{0^+}}{\frac{\ln(x)}{1/x}}}##, and, by L'Hopital, ##\displaystyle e^{\lim_{x \to{0^+}}{-x}}=1##.
This is correct.
mcastillo356 said:
Is this equal to ##f(0)##? Don't know. ##x^x=e^{x\ln(x)}##. Does this alow me to say "no", because ##\ln 0## is nonsense?.
What do you mean? "No" as an answer to what? And where do you have ##\log 0## if you use de L'Hôpital?
You could as well use a Taylor expansion of ##\log## and observe that ##y=x## goes faster to zero than ##y=\log x ## does so that the product goes to zero. Taylor will give you a quantitative measure for "faster than".

Last edited:
mcastillo356
Another try

In the exercise I'm given a function ##f(x)=x^x##. We have ##x^x:=e^{x\ln x}##, which is well defined for ##x>0##, so the domain is ##(0,\infty)##.

Now, we can define another function ##[0,\infty)\rightarrow{\mathbb R}## such that

$$g(x):=\begin{cases}{f(x)}&\text{if}& x\in{(0,\infty)}\\1 & \text{if}& x=0\end{cases}$$

Where we assume ##0^0:=1##

You can define ##g(x)## whichever you want. The question is: which properties do you require? If you do it like this, then we get a continuous extension of ##f(x)##.

mcastillo356
Hi, fresh_42

fresh_42 said:
The question is: which properties do you require?
No idea. I'm afraid I let myself struggle with maths with lots of naivety. Sooo...PF is a great, great, help, and a place where I feel at home, an island of give-and-take knowledge.

But I think I generate expectations I'm not able to tackle...

Conclusion: Thanks for helping me one more time. Much love. Forgive me if this post is improper. I felt I had to contextualize the thread.

berkeman and fresh_42

## 1. What is an exponential function?

An exponential function is a mathematical function in the form of f(x) = ab^x, where a and b are constants and x is the independent variable. It is characterized by a constant ratio between the output and input values, which creates a curved graph when plotted.

## 2. How do you calculate the critical points of an exponential function?

The critical points of an exponential function can be calculated by finding the derivative of the function and setting it equal to zero. Then, solve for the value of x that satisfies the equation. This value of x will be the critical point of the function.

## 3. Why is it important to find the critical points of an exponential function?

The critical points of an exponential function are important because they indicate where the function changes direction, from increasing to decreasing or vice versa. They also help in determining the maximum or minimum values of the function.

## 4. Can the critical points of an exponential function be negative?

Yes, the critical points of an exponential function can be negative. This means that the function has a maximum or minimum value in the negative range of the x-axis.

## 5. How can you argue the critical points of an exponential function?

The critical points of an exponential function can be argued by using the first and second derivative tests. The first derivative test helps in determining whether the critical point is a maximum or minimum, while the second derivative test confirms the concavity of the function at the critical point.

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