How do you find g(x+1) given g(x)=x32x2?

1 Answer
Feb 17, 2018

g(x+1)=(x2+2x+1)(x1)

Explanation:

A function takes in an input and returns an output.

In this function g(x), you input x and it returns x32x2. 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 (x3), subtracts two times the input squared (2x2). Putting this all together, this particular function g(x) takes in an input x and returns the expression x32x2, 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)=x32x2

so,

g(x+1)=(x+1)32(x+1)2

So now, all we have to do is use algebra to simplify this expression! We have

g(x+1)=(x+1)32(x+1)2
g(x+1)=(x+1)(x+1)(x+1)2(x2+2x+1)
g(x+1)=(x+1)(x2+2x+1)2(x2+2x+1)

Here, we see that we have a (x2+2x+1) term in common on both sides of the minus sign. We invoke the distributive property:

g(x+1)=(x2+2x+1)(x+12)
g(x+1)=(x2+2x+1)(x1)

And we're done!

SIDENOTE:

Alternatively, you could also write the answer like this:

g(x+1)=(x+1)2(x1)=(x21)(x+1)

Either way is fine.