Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

keonlo

  • 2 years ago

Is e^x the only function that has a same derivative of itself? if yes, how to proof it?

  • This Question is Open
  1. shandelman
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    f(x) = 0 has a derivative of 0.

  2. JingleBells
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes, but I think 0 is just a constant not a function, no? I believe the value of e is unique just like pi. But correct me if I'm wrong!

  3. FoolAroundMath
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\frac{df}{dx} = f\]\[\Rightarrow \frac{df}{f}=dx\]\[\Rightarrow\int \frac{df}{f} = \int dx\]\[\Rightarrow ln(f) = x+c \]\[\Rightarrow f(x) = e^{x+c}\]\[\Rightarrow f(x) = k.e^{x}\] Here, \(k\) is any real number. So, any function of the form \(f(x) = k.e^{x} \) has the same derivative as itself. (k=0 implies that f(x) = 0 also is a solution)

  4. JingleBells
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    In this case the constant k is just a translation, and the exponential still has to have e as it's base, right?

  5. FoolAroundMath
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Yes the exponential has to have \(e\) as its base, and k can be any real number..

  6. JingleBells
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Thanks!

  7. shandelman
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I would argue that f(x) = 0 is completely a function despite the fact that it is constant. It's graph passes the vertical line test, does it not? I will grant you it's not particularly the most interesting function but it is a function nonetheless.

  8. JingleBells
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well by definition it maps all points on to 0. So it is a many-to-one function. but I always wondered what x=0 on the x-y coordinates would be?

  9. shandelman
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    f(x) = 0 is a horizontal line overlapping the x-axis. x=0 is a vertical line overlapping the y-axis.

  10. JingleBells
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Does it qualify as a function was what I was wondering.

  11. FoolAroundMath
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    a function is defined by \(f: D \rightarrow R \) D is the domain R is the range such that all elements in D have a map to R and for \(x_{1}\) in D, \(f(x_{1})\) can have only one value in R. i.e. it must be one->one or many->one. f(x) = 0 satisfies this definition.

  12. alexaniandavid
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    h

  13. Not the answer you are looking for?
    Search for more explanations.

    Search OpenStudy
    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.