Inverse of function and its domain/range?

  • Thread starter Thread starter adelaide87
  • Start date Start date
  • Tags Tags
    Function Inverse
adelaide87
Messages
24
Reaction score
0

Homework Statement



a) find the inverse of f(x)= (3x+4)/(5-2x)
b) state the domain anr rage for both f and f^-1


Homework Equations





The Attempt at a Solution



for f(x)
Domain:x ≠ 5/2, x ∈ R
Range: y ≠ 3/2, y ∈ R

Inverse (my answer):

f^-1(x) = (5y-4)/(3+2y)

I want to check and see if this is correct, and also get some guidance of the domain/range for the inverse?

Thanks!
 
Physics news on Phys.org
how did you get that 3/2 is not in the range?

what about f(0.5833333333)?

f^-1 looks right though.
 
Last edited:
I evaluated the limit from x > inf and a > -infWhen you do that, the result becomes -3/2

(sorry, just realized it wasnt -'ive above. Should be -3/2!)
 
but shouldn't that be missing from the domain of f^-1 ?
The range of f is all of R and the range of f^-1 is also all of R

right?
 
What I've put above is just the domain and range for f(x)

Domain:x ≠ 5/2, x ∈ R
Range: y ≠ -3/2, y ∈ R

--------------------------------------------------------

Its the inverse, f^-1(x), that I need help finding the domain and range for now.
 
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