Linear function problem alternative solution

[Rectifier]

Hey!
I am searching for an alternative way to solve this problem
1. The problem
For a linear funtion following applies:
$f(a+1)=a+2$
Write the linear function on the $y=kx+m$ form

Homework Equations

$f(a+1)=a+2 \\ y=kx+m$

3. The attempt
Here is how I solved the problem:
$f(a+1)=k(a+1)+m \\ k(a+1)+m = a+2 \\ k(a+1)+m = a+1+1 \\ k(a+1)+m = 1 \cdot (a+1)+1$
Following gives $k=1$ and $m=1$. This means that the equation is:
$$y=x+1$$

Any ideas on how I could solve this differently?

 

[STEMucator]

It is worth noticing $f(1) = 2, \space f(2) = 3, \space f(3) = 4, ...$

You can easily find the slope in this manner:

$m = \frac{\Delta y}{\Delta x} = 1$

Then using anyone of those points, say $f(1) = 2$, you find that $y = mx + b \Rightarrow b = 1$.

 

[Ray Vickson]

Hey!
I am searching for an alternative way to solve this problem
1. The problem
For a linear funtion following applies:
$f(a+1)=a+2$
Write the linear function on the $y=kx+m$ form

Homework Equations

$f(a+1)=a+2 \\ y=kx+m$

3. The attempt
Here is how I solved the problem:
$f(a+1)=k(a+1)+m \\ k(a+1)+m = a+2 \\ k(a+1)+m = a+1+1 \\ k(a+1)+m = 1 \cdot (a+1)+1$
Following gives $k=1$ and $m=1$. This means that the equation is:
$$y=x+1$$

Any ideas on how I could solve this differently?

Much quicker: for any given $x$ let $a = x+1$. Then the equation says $f(x) = x+1$.

 

[Rectifier]

It is worth noticing $f(1) = 2, \space f(2) = 3, \space f(3) = 4, ...$

You can easily find the slope in this manner:

$m = \frac{\Delta y}{\Delta x} = 1$

Then using anyone of those points, say $f(1) = 2$, you find that $y = mx + b \Rightarrow b = 1$.

Yeah that an alternative too. Thanks for the comment :)

 

[Rectifier]

Much quicker: for any given $x$ let $a = x+1$. Then the equation says $f(x) = x+1$.

Thank you for the comment.

Do you mean like that:

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

?

If yes, what's is the next step?

 

[Ray Vickson]

Thank you for the comment.

Do you mean like that:

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

?

If yes, what's is the next step?

Sorry: there was a typo in my post. It should have said that for any given $x$, let $a = x-1$. That gives $f(x) = x+1$. Since this holds for any $x$ that's it: there is no more to be done.