Finding the Inverse of a Function with Given Conditions

  • Thread starter Thread starter Jalo
  • Start date Start date
  • Tags Tags
    Functions
Jalo
Messages
117
Reaction score
0

Homework Statement



(a) f(x) = 5x^7+6x^3+x+9. Knowing that f(-1)=-3 and that f is invertible, find f '^-1(-3)

(b) g(x) = e^x/ (e^x+1) What's g^-1(x) = ?

Homework Equations



d24cfca99a4cda57b8781f7078c57941.png

I think I'm supposed to use this equation

The Attempt at a Solution



I tried to put x in evidence, but that led me nowhere. I'm clueless in both exercises...

Thanks
 
Last edited:
Physics news on Phys.org
Your "relevant equation" is completely irrelevant because these problems say nothing about the derivative. Do you know the definition of f^{-1}? If you do the first problem at least is completely trivial. If you don't, there is no point in trying to do these until you know that definition. Look it up or talk to your teacher.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top