Difficult derivative question?

Question

It is true that Q(x)=x^5+x^3+x is a one to one function whose domain and range are all numbers.

Suppose that R is the function inverse to Q, There is no simpe algebraic way to compute values of R. compute R(3), R'(3), and R''(3).

Ok so I found the inverse but what do I do from here?

There's a hint and it says Q(R(x))=x and R(Q(x))=x. So fund an input to Q which will "output" 3. Then differentiate one of the equations, maybe more than once.

I'm really confused about what to do. If someone could explain I'd greatly appreciate it! Thanks.

Answer 1

Given:
Q(x)=x^5+x^3+x
Q(R(x))=x
R(Q(x))=x

Therefore if R(Q(x))=x;
Rinverse (R(Q(x))) = Rinverse (x)
Q(x) = Rinverse (x) = x^5+x^3+x

and if Q(R(x))=x;
Q (R(x)) = R(x)^5+R(x)^3+R(x) ; by definition of function Q
also, R(Q(x)) = R(x^5+x^3+x) = x

Find an input (or argument) for Q that will produce Q=3
By observation that would have to be x =1

Q(1) = (1)^5 + (1)^3 + 1 = 3

R(3) = R(Q(x)=3) ; since Q(x) =3, x=1
Therefore R(3) = 1 for Q(x) =3

R'(x^5+x^3+x) = R' d(x^5+x^3+x)/dx = R'(5x^4 + 3x^2 +1) = dx/dx

R' * (5x^4 + 3x^2 +1) = 1
R' = 1/(5x^4 + 3x^2 +1)
R'(3) = 1/(5(3)^4 + 3(3)^2 +1) ,evaluate

R" (5x^4 + 3x^2 +1) + R' (20x^3 + 6x) = 0
R" = - R' (20x^3 + 6x) /(5x^4 + 3x^2 +1)
R" = -(20x^3 + 6x)/(5x^4 + 3x^2 +1)^2, evaluate at x=3

Answer 2

Q(x)
y = x^5+x^3+x

R(x)
x = y^5+y^3+y

To find R(3)
3 = y^5+y^3+y
y^5+y^3+y-3 = 0
(y-1)*(y^4+y^3+2y^2+2y+3) = 0
One of the roots is 1. We are told that Q(x) is one-to-one, which means that R(x) is one-to-one as well. This means that 1 is the only root for this.
R(3) = 1.


R'(x)
x = y^5+y^3+y
1 = 5y^4*y'+3y^2*y'+y'
y' = 1/(5y^4+3y^2+1)
y'(3) = 1/(5*1^4+3*1^2+1)
= 1/9

R''(x)
1 = 5y^4*y'+3y^2*y'+y'
0 = 5y^4*y''+20y^3*y'+3y^2*y''+6y*y'+y''
-20y^3*y'-6y*y' = 5y^4*y''+3y^2*y''+y''
-2y*y'*(10y^2+3) = (5y^4+3y^2+1)*y''
y'' = -2y*y'*(10y^2+3)/(5y^4+3y^2+1)
y''(3) = -2*1*1/9*(10*1^2+3)/(5*1^4+3*1^2+1)
=-26/81

Hope I didn't make any mistakes along the way. But the process is correct.