what is the relation between $f(x+1)$ and $f(x)$?

Question

I searched so much over math sites and google but I didn't find helpful hints and required knowledge or the specific name of this topic in function.

I stuck in relation and operations on $f(x)$ which is really important for solving problems. for example relation between $f(x)$ and $f(x+1)$

in more complicated form for instance: 1. $$f(x)=3x-5$$ $$f(2x+1)=?$$

true answer is $$2f(x)+8$$

I tried this solution but it doesn't work

$f(2)$*$f(x)$+$f(1)$=?

$f(1)$=$-2$

$f(2)$=$2*3-5=1$

$f(2)$*$f(x)$+$f(1)$=?

$1(3x-5)-2$=$3x-7$ which gives us $f(x)-2$

another instance:

$$f(-x)=2f(x)+6$$ $$f(3)=?$$ true answer is $$-6$$ I tried this solution $f(x)(-1)$=$2f(x)+6$

$-f(x)$=$2f(x)+6$

$-3f(x)$=$6$

$f(x)$=$-2$

so I will be really appreciative if you tell me the relation between operation on $f(x)$ like $f({1\over x})$ , $f(x-1)$ and ..

Answer 1

Hint: For $f(x)=3x-5$. The expression $f(2x+1)$ means that you substitute $2x+1$ every time you saw $x$ in the original expression. Therefore, \begin{align*} f(2x+1)&=3(2x+1)-5\\ &=6x+3-5\\ &=6x-2. \end{align*} Since the goal is to write this in terms of $f(x)$, we try to put this expression into the form for $f(x)$, namely we see the $3x$ term in $f(x)$ while $f(2x+1)$ has a term of $6x$, which is twice $3x$. So, we expect the form to be $2f(x)=6x-10$. The problem is that the constant term doesn't line up correctly. To correct the constant term, we must add and subtract $8$ to get $$ f(2x+1)=6x-2=6x-2-8+8=6x-10+8=2f(x)+8. $$

Answer 2

One can view $f$ as a machine that takes an input (usually called $x$) and outputs something based on that input.

It is usually written in the general case as something like $f(x)=3x-5$ which says whatever our input $x$ happens to be, we output something that is (in this case) $3$ times that input and then subtract five.

In the above example it would be something like $f(\color{fuchsia}{1})=3\cdot \color{fuchsia}{1}-5$, or $f(\color{fuchsia}{2})=3\cdot \color{fuchsia}{2}-5$, or in general $f(\color{fuchsia}{x})=3\cdot \color{fuchsia}{x}-5$

In the case that the input happens to itself be written in a strange way, or even as a different function, that is okay. It is still the input, and we manipulate it in exactly the same way.

$f(\color{fuchsia}{\frac{1}{x}})=3\cdot \color{fuchsia}{\frac{1}{x}}-5$ and $f(\color{fuchsia}{x+1})=3\cdot (\color{fuchsia}{x+1})-5$ for examples.

We may choose to reorganize things afterwards at our convenience.

For example: $f(\color{fuchsia}{2x+1})=3\cdot (\color{fuchsia}{2x+1})-5=6x+3-5=6x-2$

Answer 3

There are no general way of expressing $f(x+n)$ (or $f(nx)$ or $f(\frac{x}{n})$) in terms of $f(x)$.

Functions satisfying $f(x+y)=f(x)+f(y)$ are called linear, and make a fairly small (however you want to define that) part of all functions, and make for a lot study. Similarily functions satisfying other "nice" equations can be studied.

What you call the "true answer" to the problem in the question is correct (other answers tell you how to see that), but a rather strange expression.

Answer 4

another instance:

$f(−x)=2f(x)+6 \\ f(3)=?$

For $x = 3$ we get $$ f(−3)=2f(3)+6 $$ For $x = -3$ we get $$ f(−(-3))=2f(-3)+6 \iff \\ f(3) = 2 f(-3) + 6 $$ Inserting the first into the second equation gives $$ f(3) = 2(2f(3) + 6) + 6 = 4 f(3) + 18 \iff \\ -3 f(3) = 18 \iff \\ f(3) = -6 $$

Answer 5

Substituting an equation of $ax+c$ into $f$ is one way to look at the problem, and I think you've gotten many good answers already.

However, these kinds of substitutions can more generally be considered as special cases of function composition.

In other words, [I hope] that you've seen some problems of the form $f \circ g(x)$ which is the same as $f(g(x))$. In other words, you can think of this as one function followed by another. In your case, this function is defined by $g(x)=ax+c$, and so $f \circ g(x)=f(ax+c)$. So you can think of these problems as two successive functions. There are actually a lot of ways to think about these problems, but this is the way I usually do.

Alternatively, we can consider $g \circ f(x)=a \cdot f(x)+c$, which is function composition achieved by first considering $f(x)$ and then following it by a linear transformation.

I hope that helps!