# Show that the function `f=3*2^x-2*2^-x+3x+1` has inverse. Calculate inverse in y=3

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You need to remember that a function that increases or decreases continuously, has an inverse, hence, you need to prove that the function continuously increases or decreases.

You need to evaluate derivative of the function to decide if the function increases or decreases such that:

`f'(x) = 3*2^x*ln 2 + 2*2^(-x)*ln 2 + 3`

Since the functions `2^x` and `2^(-x)` are strictly positive for all real x, hence, `3*2^x*ln 2 + 2*2^(-x)*ln 2 + 3 > 0` .

Since derivative is positive, then the function strictly increases.

You need to prove that the function is continuous over R, hence, you need to evaluate its limits over R such that:

`lim_(x->-oo) 3*2^x - 2*2^(-x) + 3x + 1 = 3*2^(-oo) - 2*2^*(oo) + 3*(-oo) + 1 => lim_(x->-oo) 3*2^x - 2*2^(-x) + 3x + 1 = 0 - oo - oo + 1 = -oo`

`lim_(x->-oo) 3*2^x - 2*2^(-x) + 3x + 1 = 3*2^(oo) - 2/(2^(oo)) + 3*oo + 1 = oo - 0 + oo + 1= oo`

Hence, the function is continuous over R.

Since the function continuously increases over R, then it has inverse.

Notice that the problem requests to evaluate `f^(-1)(x)` at `y = 3` , hence, you may evaluate x for y = 3 such that:

`y = f(x) =3*2^x - 2*2^(-x) + 3x + 1 = 3`

`3*2^x - 2*2^(-x) + 3x + 1 - 3 = 0`

`3*2^x - 2*2^(-x) + 3x -2 = 0`

`3*2^x - 2/(2^x) + 3x - 2 = 0`

`3*2^x - 2+ 3*x*2^x - 2*2^x = 0`

Hence, having x=0 as solution, you may write the equation that relates the derivative of the function and the derivative of inverse such that:

`f'(x)*(f^(-1)(x))' = 1`

Since you need to evaluate `(f^(-1)(3))' => (f^(-1)(3))' = 1/(f'(0))`

`(f^(-1)(3))' = 1/(5ln 2 + 3)`

**Hence, evaluating derivative of inverse of the function at y=3 yields `(f^(-1)(3))' = 1/(5ln 2 + 3).` **