# Thread: Investigate f(2x - 1) = x + 4

1. ## Investigate f(2x - 1) = x + 4

I ran across this question and am not sure of the answer. Needless to say, I'm having trouble with conceptualizing all of the facets of functions.

The question is: Investigate and understand f(2x - 1) = x + 4

I get two different possible answers -

1. Let z = 2x - 1
2. x = (z + 1)/2; Evaluating for x
3. f(z) = x + 4; from the original function statement
4. f(z) = [(z + 1)/2] + 4; Substituting for x from step 2
5. f(2) = [(2 + 1)/2] + 4; Evaluating with z = 2
6. f(2) = 3/2 + 4
7. f(2) = 3/2 + 8/2 = 11/2

BUT

1. Let g(x) = 2x-1
2. f(g(x)) = x + 4; Composition of 2 functions
3. g(2) = 2(2) - 1 = 3; Evaluating with g(2)
4. f(g(2)) = 3 + 4 = 7

No? What did I miss here? Am I misunderstanding a fundamental concept in functions?

2. ## Re: Investigate f(2x - 1) = x + 4

Originally Posted by B9766
I ran across this question and am not sure of the answer. Needless to say, I'm having trouble with conceptualizing all of the facets of functions.
The question is: Investigate and understand f(2x - 1) = x + 4[LIST=1][*]Let z = 2x - 1[*]x = (z + 1)/2; Evaluating for x
Not quit sure what investigate means here. What you have done in the quote is correct.
I would finish with:
because $2\left(\dfrac{z+1}{2}\right)-1=z$ then $f(z)=\dfrac{z+9}{4}$

3. ## Re: Investigate f(2x - 1) = x + 4

Plato: I'm confused by your reply. I showed 2 different approaches with 2 different answers. Both can't be correct.

I didn't follow your "finish with" response at all. The first part is obvious z =z. But it's unclear how you transformed that into the second part that is an entirely 3rd different answer.

I would really appreciate it if you would clarify this.

4. ## Re: Investigate f(2x - 1) = x + 4

Originally Posted by B9766
Plato: I'm confused by your reply. I showed 2 different approaches with 2 different answers. Both can't be correct.
I didn't follow your "finish with" response at all. The first part is obvious z =z. But it's unclear how you transformed that into the second part that is an entirely 3rd different answer.
Well, I did say that I do not understand what you are asked to do.
Here is a standard question. If $f(3x+7)=2-6x$ then what is $f(x)~?$
Solution:
Use a dummy variable, $z=3x+7$ so $x=\left(\dfrac{z-7}{3}\right)$.
That means \begin{align*}f(z)&=2-6\left(\dfrac{z-7}{3}\right)\\&=2-3(z-7)\\&=2-2z+14\\&=16-2z \end{align*}

Since $z$ is a dummy we have $f(x)=16-2x$.

5. ## Re: Investigate f(2x - 1) = x + 4

Originally Posted by B9766
Plato: I'm confused by your reply. I showed 2 different approaches with 2 different answers. Both can't be correct.

I didn't follow your "finish with" response at all. The first part is obvious z =z. But it's unclear how you transformed that into the second part that is an entirely 3rd different answer.

I would really appreciate it if you would clarify this.

$f (g (x)) \neq f (x)$

6. ## Re: Investigate f(2x - 1) = x + 4

Why are you trying to evaluate $f(2)$? You should be getting an expression for $f(x)$.

7. ## Re: Investigate f(2x - 1) = x + 4

Originally Posted by Plato
Well, I did say that I do not understand what you are asked to do.
Here is a standard question. If $f(3x+7)=2-6x$ then what is $f(x)~?$
Solution:
Use a dummy variable, $z=3x+7$ so $x=\left(\dfrac{z-7}{3}\right)$.
That means \begin{align*}f(z)&=2-6\left(\dfrac{z-7}{3}\right)\\&=2-3(z-7)\\&=2-2z+14\\&=16-2z \end{align*}

Since $z$ is a dummy we have $f(x)=16-2x$.
I followed all of this up to the last line. Why is z a "dummy" in the last line? z = 3x + 7, right?

8. ## Re: Investigate f(2x - 1) = x + 4

Originally Posted by B9766
I followed all of this up to the last line. Why is z a "dummy" in the last line? z = 3x + 7, right?
$f(z)=16-2z , f(t)=16-2t$ and $g(x)=16-2x$ are all the same function only with different names.

9. ## Re: Investigate f(2x - 1) = x + 4

I think that's a good indicator of my lack of knowledge.
I understand that f(x) = 2x says, put an input value in as x (for example 2) and get an output = 2x (or in this case f(2) = 2 * 2 = 4).
I also understand that if g(x) = 3x-1 and f(x) = 2x, then f(g(x)) = 2(3x-1) = 6x-2. Evaluating with an input of 2, f(g(2)) = 6(2)-2 = 10

I think I'm having problems wrapping my mind around a function with a polynomial on both sides of the equation.

10. ## Re: Investigate f(2x - 1) = x + 4

Thank you Plato. I understand what you're saying now. I think where my confusion arises, as I tried to relate to SlipEternal, is in reconciling the concept of a function as a program with inputs and outputs. It's easy to see f(x)=x+4 with "x+4" being the program, x being the input and f(x) being the output.

If, in order to evaluate a function in the sense of it being a program, one first has to simplify it down to a single variable as you did above, I can accept that as simply a rule without further explanation. But for the life of me I can't imagine a real-life application where it would be necessary to address a function like that.

Is this in fact the case?

11. ## Re: Investigate f(2x - 1) = x + 4

Originally Posted by B9766
I think that's a good indicator of my lack of knowledge.
I understand that f(x) = 2x says, put an input value in as x (for example 2) and get an output = 2x (or in this case f(2) = 2 * 2 = 4).
I also understand that if g(x) = 3x-1 and f(x) = 2x, then f(g(x)) = 2(3x-1) = 6x-2. Evaluating with an input of 2, f(g(2)) = 6(2)-2 = 10

I think I'm having problems wrapping my mind around a function with a polynomial on both sides of the equation.
Suppose $f(x)=\color{red}{x^2-x}~\&~g(x)=\color{blue}{2x+1}$
Then $f\circ g(t)=f(g(t))=\color{red}{(}2x+1\color{red}{)^2-(}2x+1\color{red}{)}$ and $g\circ f(t)=g(f(t))=2(x^2-x)+1$

12. ## Re: Investigate f(2x - 1) = x + 4

Yes Plato. I'm not sure where the t's came from but am guessing they're typos. I think you meant to say:
Suppose f(x)=x^2−x & g(x)=2x+1
Then f∘g(x)=f(g(x))=(2x+1)^2−(2x+1) and g∘f(x)=g(f(x))=2(x^2−x)+1

But in both cases the functions were expressed a monomials - f(x) and g(x). Where I get confused is when the function is expressed as a polynomial like f(x^2-5) or g(3x+4). In these cases, the concept of plugging a value in for the variable is not obvious in the way it's done nor in the outcome.

If I have a table of x and y values where y = f(x) = 2x, then I know that plugging in the value x=5 will produce the output y = 10.

If my function is f(x-4) = x^2, then, is my table of values expressed in terms of x-4 and y or x and y? Let's say I make x = 8, do I get f(8-4) = 8^2? I know that's wrong but somehow we violated the analogy of a computer with inputs and outputs.

Part of this is that I can fathom why the function f(x) = x^2+4x+2 would be used in real life. I can't imagine why anyone would want to evaluate something like:
f(x^4-6x^3+5x^2-12x+7) = 36x^2+60x-199

Is there anything in the world of calculus or physics that would require such an evaluation?

I'm sure I'm belaboring the point. But I really need to understand how this works because I know I'll need it in calculus.