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

1 Answer
Oct 7, 2017

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

#w(x+1)=(x+1)^3-2(x+1)#

Explanation:

You have:
#w(x)=x^3-2x#

If you wanted to find #w(u)#, all you need to do is sub in #u# wherever #x# is.

So #w(u)=u^3-2u#

Similarly, if you wanted to find #w(x+1)#, you just sub in #x+1# wherever #x# is.

So #w(x+1)=(x+1)^3-2(x+1)#

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