How do you find the inverse of #f(x)= x^5+x^3+x#?

In general quintic equations are not solvable by radicals.

The inverse of #f(x) = x^3+x# is tricky enough:

Using Cardano's method, we can find that

#f^-1(y) = root(3)((9y+sqrt(3y^2+12))/18) + root(3)((9y-sqrt(3y^2+12))/18)#

We can show that #g(x) = x^5+x^3+x# does have an inverse, in that it is continuous, monotonically increasing, with derivative always #>= 1#.

#d/(dx)g(x) = 5x^4+3x^2+1 >= 1# for all #x in RR#

since #5x^4 >= 0# and #3x^2 >= 0# for all #x in RR#

You can define:

#g^-1(y) = x in RR : x^5+x^3+x = y#

Update

In theory, it's possible to find a Tschirnhaus transformation to bring #x^5+x^3+x-y = 0# into Bring-Jerrard normal form:

#x^5+k_0x+k_1 = 0#

where #k_0# and #k_1# would of course depend on the value of #y#.

Then you could express the solution in terms of the Bring radical.

In practice, this is way too cumbersome, does not really gain you much and you would be better off just approximating numerically.

In any case, this is beyond the scope of Precalculus level.

Given:

#f(x) = x^5+x^3+x#

We have:

#f'(x) = 5x^4 + 3x^2 + 1 >= 1# for all #x in RR#

So #f(x)# is continuous and strictly monotonically increasing on #RR#.

#lim_(x->-oo) f(x) = -oo# and #lim_(x->oo) f(x) = oo#

So #f(x)# is a bijection between #RR# and #RR#, with a well defined, strictly monotonically increasing inverse #f^(-1)(y)#.

We can define #f^(-1)(y)# as:

#f^(-1)(y) = x in RR : x^5+x^3+x = y#

The right hand side means that #f^(-1)(y)# is #x in RR# such that #x^5+x^3+x = y#. We have already found that such a value is defined and unique for each #y in RR#, so #f^(-1)(y)# is well defined.

For a given #y#, the equation:

#x^5+x^3+x-y = 0#

generally has no solution expressible in terms of elementary functions. For some values it does have a simple solution, for example if #y=3# then #x=1#, but generally it does not.

We can find numerical approximations using Newton's method:

For an initial approximation we can use:

#a_0 = root(5)(y)#

Then iterate to get better approximations using the formula:

#a_(i+1) = a_i - (f(a_i))/(f'(a_i)) = a_i - (a_i^5+a_i^3+a_i - y)/(5a_i^4 + 3a_i^2+1)#

This will converge fairly rapidly for any value of #y#.

Here are the graphs of #f(x)#, #f^(-1)(y)# and #y=x#. Note that the inverse is simply the reflection of the original function in the line #y=x#...

graph{(y-x)(y-(x^5+x^3+x))(x-(y^5+y^3+y)) = 0 [-5.143, 4.857, -2.38, 2.62]}