-1

So the first thing I found is if the function is one-to-one, because we know that if it is there is a inverse function of that function:

I found the derivative of the function: $f'(x)=2e^{2x}+12x^2$ where $f'(x) > 0$ for $x \in (-∞,∞)$ which implies that the function is one-to-one.

We have then per definition that the inverse function is given by: $$y = f^{-1}(x) \impliedby x = f(y)$$

so we define that $x = f(y) = e^{2y}+4y^3 \implies \ln(x) = 2y\ln(e)+ 3\ln(4y)$

I dont find a way to solve this equation for $y$. Have I done this correctly or?

Sorry for any grammar or spelling mistakes, english is not my first language. Thanks for any help!

Gary
  • 31,845
SliVetle
  • 1
  • Just because a function is elementary it does not mean its inverse is elementary. Look up the Lambert $W$-function for instance. – Gary Mar 14 '23 at 10:22
  • You have made a mistake in your manipulation since $\ln (a+b)$ does not simplify, you assumed $\ln(e^{2y}+4x^3) = \ln,e^{2y}+\ln(4x^3)$ which it does not. remember $\ln(ab)=\ln , a + \ln , b$ – asymptotic Mar 14 '23 at 10:26
  • You recieved 2 answers to your question. Is any of them what you needed? If so, consider accepting the best answer and upvoting all useful answers you got. That's how the site works. – 5xum Mar 16 '23 at 12:20

2 Answers2

1

$$x = f(y) = e^{2y}+4y^3 \implies \ln(x) = 2y\ln(e)+ 3\ln(4y)$$

False. It is not true, in general, that $\ln(a+b)=\ln a + \ln b$. In fact, what $\ln a + \ln b$ is equal to $\ln ab$, not $\ln(a+b)$.

In general, inverses of functions such as yours are rarely elementary functions, which means it is unlikely that there exists a "nice" way of writing the inverse of your $f$.

5xum
  • 123,496
  • 6
  • 128
  • 204
0

As far as I can judge, the following equation has no analytical solution:

$$e^x+x=n$$

So, your equation $e^{2x} + 4x^3$, which is more complicated than yours, can't be inversed.

Dominique
  • 2,130