Question

I am confused what to do when I get to the 1 times g^2(1)-w(5)

Answer 1

The answer is 34.

To solve this, let's rewrite M(x) so it's easier:

$M(x)=g^2(x)-w(x+4)\Rightarrow M(x)=g(x)\cdot g(x)-w(x+4)$

Now we can replace x = 1:

$M(1)=g(1)\cdot g(1)-w(1+4)$

The next thing to do is look at the table. What's the value of g(x) when x = 1?

In the table we can see that g(1) = -6

and since w(1 + 4) = w(5), we can llok in the table the value of w(5) = 2

All we need to do is replace these values:

$M(1)=g(1)\cdot g(1)-w(5)\Rightarrow M(1)=(-6)\cdot(-6)-2=36-2=34$