A function takes in an input and returns an output.
In this function #g(x)#, you input #x# and it returns #x^3-2x^2#. We can also look at this as this: you input #x# (what's enclosed by the parentheses) and this function #g(x)# takes the input (x), cubes it (#x^3#), subtracts two times the input squared (#-2x^2#). Putting this all together, this particular function #g(x)# takes in an input #x# and returns the expression #x^3-2x^2#, with #x# representing the input (which could be anything).
Cool! So what exactly does the question mean by #g(x+1)#? What this means is that the entire expression #x+1# is the input! So, because
#g(x)=x^3-2x^2#
so,
#g(x+1)=(x+1)^3-2(x+1)^2#
So now, all we have to do is use algebra to simplify this expression! We have
#g(x+1)=(x+1)^3-2(x+1)^2#
#g(x+1)=(x+1)(x+1)(x+1)-2(x^2+2x+1)#
#g(x+1)=(x+1)(x^2+2x+1)-2(x^2+2x+1)#
Here, we see that we have a #(x^2+2x+1)# term in common on both sides of the minus sign. We invoke the distributive property:
#g(x+1)=(x^2+2x+1)(x+1-2)#
#g(x+1)=(x^2+2x+1)(x-1)#
And we're done!
SIDENOTE:
Alternatively, you could also write the answer like this:
#g(x+1)=(x+1)^2(x-1)=(x^2-1)(x+1)#
Either way is fine.
