# If `f(x) = x^3 + x` and `g(x) = f^(-1)(x),` what is the value of `g'(2)` ?

*print*Print*list*Cite

Hi, ntretheway,

Let's see. Well, if f(x) and g(x) are inverses, then:

**f(g(x)) = x**

Taking thexderivatives of each side, we have:

**f'(g(x)) * g'(x) = 1**

Then, dividing by f'(g(x)) on each side:

**g'(x) = 1/f'(g(x))**

So:

**g'(2) = 1/f'(g(2))**

So, we need to find **g(2)**. But, if f(x) and g(x) are inverses, then, that means we need **to find x where f(x) = 2**, since **g(2) = x**. So:

**f(x) = x^3 + x = 2**

To solve for x, we can, **by inspection, see that x = 1**. So:

**g(2) = 1**

So, we are looking for:

g'(2) = 1/f'(g(2))

**g'(2) = 1/f'(1)**

So, now, we need the derivative of f(x):

**f'(x) = 3x^2 + 1**

f'(1) = 3*1^2 + 1

**f'(1) = 4**

So, plugging that into the equation

g'(2) = 1/f'(1)

**g'(2) = 1/4**

Good luck, ntretheway. I hope this helps.

I wasn't sure if this would be squeezing too much into one post, so I'll make a separate one. This can also be seen graphically by graphing a function and its inverse, then looking at tangent lines. I'll have to graph a different function since `g` doesn't look to be particularly nice.

Here are inverse functions (thus the reflection across the green dashed line `y=x`) and their tangent lines. Because of this reflection, the slope of the black tangent line at `(1,2)` is the reciprocal of the slope of the red tangent line at `(2,1).` Exactly what we found above.

This isn't a completely rigorous proof, but it captures the idea of the general proof of the theorem relating the derivatives of inverse functions, and makes it easy to see why the theorem is true.

By definition,

`g'(2)=lim_(x->2)(g(2)-g(x))/(2-x).`

Note that since `f(1)=2,` we have `g(2)=g(f(1))=1,` since `g=f^(-1).` Also, since `f` is one-to-one and onto, any `x` can be written as `f(t)` for exactly one value `t.` Making these replacements, we get

`(g(2)-g(x))/(2-x)=(1-g(f(t)))/(f(1)-f(t))=(1-t)/(f(1)-f(t))=((f(1)-f(t))/(1-t))^(-1).`

Remembering the relationship between `x` and `t,` and relying on continuity, we see that as `x->2,` `t->1.` Using a basic fact about limits and reciprocals, we see that

`g'(2)=lim_(x->2)(g(2)-g(x))/(2-x)=\lim_(t->1)((f(1)-f(t))/(1-t))^(-1)=1/(f'(1)).`

Since `f'(1)=3(1)^2+1,`

**we get** `g'(2)=1/4.`