# Thread: finding derivative of an inverse function given f(x) = x^5 +3x +6

1. ## finding derivative of an inverse function given f(x) = x^5 +3x +6

given that f(x) = x^5 +3x +6 is one to one, find (f^-1)' (10) (the derivative of the inverse function at x=10)

I know that the derivative of an inverse function = 1/f'(f^-1 (x))
f'(x) = 5x^4+3

However, I am not sure about how to find the value of f^-1 (10)

i know that point (10, y) on the inverse is the point (y, 10) on f(x)
So if I can isolate for the x value in the function f(x) = x^5 +3x +6 I will know (10, y) and I will know what f^-1 (10) is

however, f(x) is a quintic function, and I don't know how to isolate for the x value.

I've concluded there must be an easier way to solve this that I am not seeing, so any guidance would be much appreciated.

2. I'm assuming you would solve for the derivative as normal.

Then since its f prime to the negative 1 power you would just change it to 1 / f prime. Then solve for your point at x = 10

If its f to the negative 1 prime.. then invert and use the quotient rule.

3. Originally Posted by fresh
given that f(x) = x^5 +3x +6 is one to one, find (f^-1)' (10) (the derivative of the inverse function at x=10)

I know that the derivative of an inverse function = 1/f'(f^-1 (x))
f'(x) = 5x^4+3

However, I am not sure about how to find the value of f^-1 (10)

i know that point (10, y) on the inverse is the point (y, 10) on f(x)
So if I can isolate for the x value in the function f(x) = x^5 +3x +6 I will know (10, y) and I will know what f^-1 (10) is

however, f(x) is a quintic function, and I don't know how to isolate for the x value.

I've concluded there must be an easier way to solve this that I am not seeing, so any guidance would be much appreciated.
I cannot tell you a general approach to solving quintics, but the integer root theorem lets you solve this one with ease.

$x^5 + 3x + 6 = 10 \iff x^5 + 3x - 4 = 0.$

Now, by the integer root theorem, if there is a rational root, it will be an integer that divides 4 evenly. There are only six possibilities, - 4, - 2, - 1, 1, 2, and 4. Obviously, a negative number will not work. Just as obviously, 2 and 4 won't work. How about + 1.

$1^5 + 3 * 1 + 6 = 1 + 3 + 6 = 10.$

$\therefore f(1) = 10 .$

4. ## Translate the graph at (1,10) to (0,0)

P(x) = x^5+3x+6
P(x+1)-10 = (x+1)^5+3(x+1)+6-10
P(x+1)-10 = x^5+5x^4+10x^3+10x^2+8x
P(x+1)-10 ~ 8x near the origin

P(x+1)-10 at (0,0) has the same slope as P(x) at (1,10)
P'(1) = 8
InvP(10) = -1/8

5. Originally Posted by Bob Brown MSEE
P(x) = x^5+3x+6
P(x+1)-10 = (x+1)^5+3(x+1)+6-10
P(x+1)-10 = x^5+5x^4+10x^3+10x^2+8x
P(x+1)-10 ~ 8x near the origin

P(x+1)-10 at (0,0) has the same slope as P(x) at (1,10)
P'(1) = 8
InvP(10) = -1/8
I do not understand this answer.

What is the relevance of P(x + 1)? There is no difficulty in figuring out the derivative of P(x).

$P'(x) = 5x^4 + 3 \implies P'(1) = 8 \implies \dfrac{1}{P'(1)} = \dfrac{1}{8}.$

The question asked was how to find $P^{-1}(10).$

Once you know that $P^{-1}(10) = 1,$ the problem is trivial.

Clearly I am missing something subtle.

6. Originally Posted by JeffM
I cannot tell you a general approach to solving quintics, but the integer root theorem lets you solve this one with ease.

$x^5 + 3x + 6 = 10 \iff x^5 + 3x - 4 = 0.$

Now, by the integer root theorem, if there is a rational root, it will be an integer that divides 4 evenly. There are only six possibilities, - 4, - 2, - 1, 1, 2, and 4. Obviously, a negative number will not work. Just as obviously, 2 and 4 won't work. How about + 1.

$1^5 + 3 * 1 + 6 = 1 + 3 + 6 = 10.$

$\therefore f(1) = 10 .$
Thank you! the problem is really quite simple with the integer root theorem applied.

7. ## Jeff

Originally Posted by JeffM

The question asked was how to find $P^{-1}(10).$

Once you know that $P^{-1}(10) = 1,$ the problem is trivial.

Clearly I am missing something subtle.
Hi Jeff,

Its good to talk with you again, I've been away.
On this post, (my bad) I miss read the OP and had the impression that he had guessed by inspection that (1,10) was on the graph. If so, then ...
STEPS:
1) Show that (1, 10) is on the polynomial.
2) Calculate the slope at (1,10)

My solution was simply a way to do both steps at once.
HOWEVER I went back to re-read the post -- he did not have $P^{-1}(10).$ -- Clearly you did not miss something subtle

Thanks for catching my error.

8. Bob

Ahh I see. Not a problem. I really did think there might have been something subtle going on.

Thanks for following up.