Reverse Composition to find composition of a derivative

In summary: I don't see any terribly useful ways to simplify those. Why do you think you have to simplify?You could factor a (x-1) out of the terms in the radical in the denominator of the second form but that really doesn't look all that much simpler...In summary, the problem involves finding the derivative of a function g(x) given that f(x+1)=√(x^2-2x) and g(x)=f(√x). The steps to find g'(x-1) involve substituting x with x+k to find f(x+k) and then determining the value of k that makes f(x+k)=f(x). Once f(x) is found, it is substituted into the equation g
  • #1
ninfinity
9
0

Homework Statement


[tex]
I\!f~ f(x+1)=\sqrt{x^2-2x}~~~and~~~g(x)=f(\!\sqrt{x})
[/tex]
[tex]
Find: g'(x-1)
[/tex]

Homework Equations


In order to find [itex]g'(x-1)[/itex] I know the following steps have to be taken:
[tex]
f(x+1) \rightarrow f(x) \rightarrow f(\sqrt{x}) = g(x) \rightarrow g'(x) \rightarrow g'(x-1)
[/tex]

The Attempt at a Solution


Composing ##\sqrt{x}## into ##f(x)## isn't hard, given that I have ##f(x)##. Finding the derivative and then composing ##x-1## into that can also be done without much difficulty.
The only problem that I find myself coming across is how exactly to move from ##f(x+1)## to ##f(x)##. I don't remember any sort of formal equation of something like this, so I started just playing around with it.
I defined ##h(x)=x+1## so that my original equation becomes ##f(h(x))= \sqrt{x^2-2x}##.
Then I tried to logic out what had to be done to ##h(x)## to become ##f(h(x))##. First I decided it needed to be squared. $$h(x)^2 = (x+1)^2 = x^2+2x+1$$It also needs to be taken to a one-half power, but doing that would negate the square, and so something else needs to be happening under the radical.
So I tried $$\sqrt{h(x)^2-1}=\sqrt{(x+1)^2-1}=\sqrt{x^2+2x+1-1}=\sqrt{x^2+2x}$$
Unfortunately, that does not equal ##f(x+1)##.
I happen to be stuck here and cannot figure out how to move forward.
 
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  • #2
ninfinity said:

Homework Statement


[tex]
I\!f~ f(x+1)=\sqrt{x^2-2x}~~~and~~~g(x)=f(\!\sqrt{x})
[/tex]
[tex]
Find: g'(x-1)
[/tex]

Homework Equations


In order to find [itex]g'(x-1)[/itex] I know the following steps have to be taken:
[tex]
f(x+1) \rightarrow f(x) \rightarrow f(\sqrt{x}) = g(x) \rightarrow g'(x) \rightarrow g'(x-1)
[/tex]

The Attempt at a Solution


Composing ##\sqrt{x}## into ##f(x)## isn't hard, given that I have ##f(x)##. Finding the derivative and then composing ##x-1## into that can also be done without much difficulty.
The only problem that I find myself coming across is how exactly to move from ##f(x+1)## to ##f(x)##. I don't remember any sort of formal equation of something like this, so I started just playing around with it.
I defined ##h(x)=x+1## so that my original equation becomes ##f(h(x))= \sqrt{x^2-2x}##.
Then I tried to logic out what had to be done to ##h(x)## to become ##f(h(x))##. First I decided it needed to be squared. $$h(x)^2 = (x+1)^2 = x^2+2x+1$$It also needs to be taken to a one-half power, but doing that would negate the square, and so something else needs to be happening under the radical.
So I tried $$\sqrt{h(x)^2-1}=\sqrt{(x+1)^2-1}=\sqrt{x^2+2x+1-1}=\sqrt{x^2+2x}$$
Unfortunately, that does not equal ##f(x+1)##.
I happen to be stuck here and cannot figure out how to move forward.

You seem to be over-thinking on the problem. The first task as you said is to find f(x).

You have f(x+1). Replace x with something which can give you f(x).
 
  • #3
Pranav-Arora said:
You have f(x+1). Replace x with something which can give you f(x).

I don't understand. Replace x with some arbitrary coefficient?
 
  • #4
ninfinity said:
I don't understand. Replace x with some arbitrary coefficient?

No no. Replace x with something like x+k, which turns x+1 to x.
 
  • #5
Pranav-Arora said:
No no. Replace x with something like x+k, which turns x+1 to x.

So...it would be something along the lines of f(x+k)=√x2-2x? I'm not making the connection.
 
  • #6
ninfinity said:
So...it would be something along the lines of f(x+k)=√x2-2x? I'm not making the connection.

Replace x with x+k, you get ##f(x+k+1)=\sqrt{(x+k)^2-2(x+k)}##. What should be k if LHS is to be ##f(x)##?
 
  • #7
I'll admit I was still a little confused by what you meant, so I decided to try out a few numbers first. At last I tried k=-1 and noticed that inside of the function, f(x+k+1) = f(x) if k=-1, are things like this true for all composition functions and the like?
 
  • #8
This is obviously a calculus problem, so belongs in the Calculus & Beyond section. I am moving it there.
 
  • #9
ninfinity said:
I'll admit I was still a little confused by what you meant, so I decided to try out a few numbers first. At last I tried k=-1 and noticed that inside of the function, f(x+k+1) = f(x) if k=-1
Yes, k=-1 is the right choice. So what is ##f(x)## now?

are things like this true for all composition functions and the like?
Yes, they are.
 
  • #10
Mark44 said:
This is obviously a calculus problem, so belongs in the Calculus & Beyond section. I am moving it there.

I apologize for that, since the question was focusing on the algebra of the problem I thought it was best suited for the precalculus section.

Pranav-Arora said:
Yes, k=-1 is the right choice. So what is f(x) now?

$$f(x)=\sqrt{x^2-4x+3}$$
 
  • #11
ninfinity said:
$$f(x)=\sqrt{x^2-4x+3}$$

That's correct. You have f(x) now, do the remaining part.
 
  • #12
Pranav-Arora said:
That's correct. You have f(x) now, do the remaining part.

Then it follows that
$$f(\sqrt{x})=\sqrt{x-4\sqrt{x}+3}=g(x)$$
$$g'(x)=\frac{\sqrt{x}-2}{2\sqrt{x^2-4\sqrt{x^3}+3x}}$$
$$g'(x-1)=\frac{\sqrt{x-1}-2}{2\sqrt{(x-1)(x+2)-4\sqrt{(x-1)^3}}}$$

Those last two functions need to be simplified still further . The problem I am having, however, is that each time I go about this I seem to end up going in circles. I've been thinking about trying to factor using fractional exponents, but I'm not sure if that would get me somewhere useful. Any advice?
 
  • #13
ninfinity said:
Then it follows that
$$f(\sqrt{x})=\sqrt{x-4\sqrt{x}+3}=g(x)$$
$$g'(x)=\frac{\sqrt{x}-2}{2\sqrt{x^2-4\sqrt{x^3}+3x}}$$
$$g'(x-1)=\frac{\sqrt{x-1}-2}{2\sqrt{(x-1)(x+2)-4\sqrt{(x-1)^3}}}$$

Those last two functions need to be simplified still further . The problem I am having, however, is that each time I go about this I seem to end up going in circles. I've been thinking about trying to factor using fractional exponents, but I'm not sure if that would get me somewhere useful. Any advice?

I don't see any terribly useful ways to simplify those. Why do you think you have to simplify?
You could factor a (x-1) out of the terms in the radical in the denominator of the second form but that really doesn't look all that much simpler.
 
Last edited:

1. What is Reverse Composition to find composition of a derivative?

Reverse Composition is a mathematical concept used in calculus to find the composition of a derivative function. It involves taking the derivative of a given function and then finding the original function by "reversing" the derivative operation.

2. How is Reverse Composition used in calculus?

Reverse Composition is used in calculus to find the original function when only the derivative function is known. This can be useful in solving complex problems involving derivatives, such as optimization and related rates.

3. Can Reverse Composition be used for any type of function?

Yes, Reverse Composition can be used for any continuous function, as long as the derivative of the function exists. This includes polynomial, exponential, logarithmic, and trigonometric functions.

4. What are the steps for using Reverse Composition to find the composition of a derivative?

The steps for using Reverse Composition are as follows: 1) Take the derivative of the given function, 2) Set the derivative equal to the derivative of the inner function multiplied by the derivative of the outer function, 3) Solve for the inner function, 4) Use the inverse of the derivative operation to find the original function.

5. What is the purpose of using Reverse Composition instead of just taking the derivative?

Reverse Composition allows us to find the original function when only the derivative is known. This can be helpful in solving real-world problems where we may only have information about the rate of change of a function, rather than the function itself.

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