SOLUTION: if f(x)=(2-xsquare) how to solve : f(x+h)-f(x)

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: if f(x)=(2-xsquare) how to solve : f(x+h)-f(x)       Log On


   



Question 94843: if f(x)=(2-xsquare)
how to solve :
f(x+h)-f(x)

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
.
f%28x%29+=+2+-+x%5E2 <=== call this equation 1
.
Next you need to find f%28x%2Bh%29. What this means is you return to f%28x%29+=+2+-+x%5E2 and
you replace every x in that equation with x%2Bh. When you do that replacement you get:
.
f%28x%2Bh%29+=+2+-+%28x%2Bh%29%5E2 <=== call this equation 2
.
To simplify this you need to square the x%2Bh on the right side. Do this by multiplying:
.
%28x%2Bh%29%5E2+=+%28x%2Bh%29%2A%28x%2Bh%29+=+x%5E2+%2B2xh+%2B+h%5E2
.
Substitute x%5E2+%2B2xh+%2B+h%5E2 into equation 2 in place of %28x%2Bh%29%5E2 and equation 2 becomes:
.
f%28x%2Bh%29+=+2+-+%28x%5E2+%2B+2xh+%2B+h%5E2%29+=+2+-+x%5E2+-+2xh+-+h%5E2 <=== call this equation 3
.
The problem asked you to find f%28x%2Bh%29+-+f%28x%29. From equation 3 we know that:
.
f%28x%2Bh%29+=+2+-+x%5E2+-+2xh+-+h%5E2
.
and from equation 1 we know:
.
f%28x%29+=+2+-+x%5E2
.
So we substitute the right side of these two equations into f%28x%2Bh%29+-+f%28x%29 in place
of f%28x%2Bh%29 and f%28x%29 to get:
.
f%28x%2Bh%29+-+f%28x%29+=+2+-+x%5E2+-+2xh+-+h%5E2+-+%282+-+x%5E2%29
.
On the right side, since the set of parentheses are preceded by a minus sign, the parentheses
can be removed if we change the signs of the two terms in the parentheses. This makes the
equation become:
.
f%28x%2Bh%29+-+f%28x%29+=+2+-+x%5E2+-+2xh+-+h%5E2+-+2+%2B+x%5E2%29
.
Now on the right side you can notice that the +2 and the -2 cancel each other as does the
-x%5E2+ and the x%5E2. Those terms, therefore, disappear and you are left with:
.
f%28x%2Bh%29+-+f%28x%29+=+-+2xh+-+h%5E2
.
Hope this helps you to understand the problem a little more.
.