Find the inverse of the polynomial.

[jrjack]

Homework Statement

Find $(f^{-1})'(a)$ of: $f(x)=\sqrt{x^{3}+x^{2}+x+22}$ ; a=5.

Homework Equations

$$(f^{-1})'(a)=\frac{1}{f'((f^{-1})(a))}$$

The Attempt at a Solution

Well, I know to find an inverse: I need to set the equation equal to y, solve for x, then swap x and y. But I don't know how to solve this for x.
I have been searching the internet for a lesson on this type of problem but I am not finding anything.
I can find the derivative, but I don't think that helps me here.
$$f'(x)=\frac{1}{2}({x^{3}+x^{2}+x+22})^{-\frac{1}{2}}(3x^{2}+2x+1)$$

 

[Dick]

Homework Statement

Find $(f^{-1})'(a)$ of: $f(x)=\sqrt{x^{3}+x^{2}+x+22}$ ; a=5.

Homework Equations

$$(f^{-1})(a)=\frac{1}{f'((f^{-1})(a))}$$

The Attempt at a Solution

Well, I know to find an inverse: I need to set the equation equal to y, solve for x, then swap x and y. But I don't know how to solve this for x.
I have been searching the internet for a lesson on this type of problem but I am not finding anything.
I can find the derivative, but I don't think that helps me here.
$$f'(x)=\frac{1}{2}({x^{3}+x^{2}+x+22})^{-\frac{1}{2}}(3x^{2}+2x+1)$$

Don't try to find a general formula for the inverse. It's way too hard. You only need the value of the inverse at a=5. What's f^(-1)(5)? It's perfectly ok to guess.

 

[jrjack]

Well, $$f(5)=\sqrt{177}\approx13.304$$
So, is the inverse 1/13.304, but then what is the derivative of the inverse?

Or the inverse of the derivative? at a=5$$f'(x)=\frac{1}{2}({x^{3}+x^{2}+x+22})^{-\frac{1}{2}}(3x^{2}+2x+1)$$
$$f'(5)=\frac{66}{2\sqrt{177}}$$
$$(f^{-1})'(5)=\frac{2\sqrt{177}}{66}\approx\frac{13}{33}$$
I'm not really sure of the correct guessing process.

 

[jrjack]

My choices were 2/3, 5/3, 4/3, or 7/3. I guessed 2/3, but was wrong and the correct answer was 5/3. My guessing needs work.

 

[Dick]

Well, $$f(5)=\sqrt{177}\approx13.304$$
So, is the inverse 1/13.304, but then what is the derivative of the inverse?

Or the inverse of the derivative? at a=5$$f'(x)=\frac{1}{2}({x^{3}+x^{2}+x+22})^{-\frac{1}{2}}(3x^{2}+2x+1)$$
$$f'(5)=\frac{66}{2\sqrt{177}}$$
$$(f^{-1})'(5)=\frac{2\sqrt{177}}{66}\approx\frac{13}{33}$$
I'm not really sure of the correct guessing process.

To find f^(-1)(5) you want to find a value of x such that f(x)=5. As you said, f(5)=sqrt(177). So that's not it. f(0)=sqrt(22). That's not 5 either. Put some more values into f(x) and see if you can get it to come out 5.

 

[jrjack]

f(1)=5

 

[Dick]

f(1)=5

Ok, so f^(-1)(5)=1. That's the f^(-1)(a) in your formula, right?

 

[jrjack]

In this formula?$$(f^{-1})(a)=\frac{1}{f'((f^{-1})(a))}$$
$$(f^{-1})(a)=\frac{1}{f'(1)}$$
then
$$f'(1)=\frac{1}{2}(25)^{-\frac{1}{2}}(6)$$
equals 3/5, and the inverse is 5/3.

So, is this the way I should go about these kind of problems?

 

[Dick]

In this formula?$$(f^{-1})(a)=\frac{1}{f'((f^{-1})(a))}$$
$$(f^{-1})(a)=\frac{1}{f'(1)}$$
then
$$f'(1)=\frac{1}{2}(25)^{-\frac{1}{2}}(6)$$
equals 3/5, and the inverse is 5/3.

So, is this the way I should go about these kind of problems?

Yes. That's the way to go. If a function is hard or impossible to invert, they'll probably make it easy to guess a specific value of the inverse, like in this problem.

BTW your formula should read (you're missing a prime)

$$(f^{-1})'(a)=\frac{1}{f'((f^{-1})(a))}$$

 

[jrjack]

I am still a little confused, but thank you very much for your help.

I need to find and practice another one of these.