# f(x) = x^3 + x Derive the inverse of f(x) f^-1(x)

### 2 Answers | Add Yours

I suspect that you have misunderstood the problem. To find the inverse of `y=x^3+x` you would exchange x and y and then solve for y; so solve `y^3+y=x` for y in terms of x. One way involves a substitution using the hyperbolic sin (sinh).

The solution (from an online calculator) is:

`f^(-1)(x)=root(3)(2/3)/root(3)(sqrt(3)sqrt(27x^2+4)-9x)-root(3)(sqrt(3)sqrt(27x^2+4)-9x)/(root(3)(2)3^(2/3))`

I suspect you were actually asked to find something like the value of `(f^(-1))'(2)` (find the derivative of the inverse function at a given point.) This you can do without calculating the inverse of the function using the derivative of an inverse function theorem from calculus.

If f is differentiable on some interval and f is invertible with inverse function g, then `g'(x)=1/(f'(g(x)))`

To find `(f^(-1))'(2)` we get `(f^(-1))'(2)=1/(f'(f^(-1)(2)))`

Since f(1)=2, we know that `g(2)=f^(-1)(2)=1` so we get

`(f^(-1))'(2)=1/(f'(1))`

`f'(x)=3x^2+1==>f'(x)=4` so finally

`(f^(-1))'(2)=1/4`

** The solution above is incorrect. It is easily checked by graphing: the graph of the inverse is a reflection of the function over the line y=x.

Further, `f'(x)*(f^(-1))'(x) !=1` in general. (Try `y=x^2, x>=0` and its inverse `y=sqrt(x)` . However, if (a,b) is on f and (b,a) is on its inverse, then the slope of the tangent line on f at a is the reciprocal of teh slope of the tangent line at b on the inverse.

You should use the following relation between derivative of a function and its inverse, such that:

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

Hence, you may find `(f^(-1)(x))'` dividing the equation above by `f'(x),` such that:

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

You need to differentiate the function `f(x)` with respect to `x` , such that:

`f'(x) = (x^3+x)' => f'(x) = 3x^2 + 1`

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

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