How do you find w(x+1) given w(x)=x^3-2x?

Oct 7, 2017

You just sub in $x + 1$ wherever $x$ is.

$w \left(x + 1\right) = {\left(x + 1\right)}^{3} - 2 \left(x + 1\right)$

Explanation:

You have:
$w \left(x\right) = {x}^{3} - 2 x$

If you wanted to find $w \left(u\right)$, all you need to do is sub in $u$ wherever $x$ is.

So $w \left(u\right) = {u}^{3} - 2 u$

Similarly, if you wanted to find $w \left(x + 1\right)$, you just sub in $x + 1$ wherever $x$ is.

So $w \left(x + 1\right) = {\left(x + 1\right)}^{3} - 2 \left(x + 1\right)$

If the question also wants you to then expand and simplify this, you could expand the expression ${\left(x + 1\right)}^{3} - 2 \left(x + 1\right)$, and then add/subtract common terms.