Working with a difference quotient involving a square root
Working with a difference quotient involving a square root
Suppose f (x) = x and suppose we want to simplify the differnce quotient
f (x + h) - f (x) h
as much as possible (say, to eliminate the h in the denominator).
Substituting the definition of f into the quotient, we have
f (x + h) - f (x) x + h - x
=
h
h
at which point we are stuck, as far as basic algebraic manipulations go.
To the rescue, however, comes the conjugate.
For any expression of the form A- B, we say its conjugate is A+ B, and vice versa: the conjugate of
the latter is the former: we get to the expressions conjugate by simply changing the sign of the operation
between the two square root expressions (plus to minus, or minus to plus).
By writing the number 1 as the expression's conjugate divided by itself, we get a powerful tool for manipulating these types of expressions.
With
x+h- x
,
h
the conjugate we want to use is x + h + x, so we multiply our expression by the conjugate over itself:
x+h-
x =
x+h-
x
?
x
+
h
+
x .
h
h
x+h+ x
The key idea is that the numerators multiply in a nice way. Note that the two numerators together have the form
(A - B) ? (A + B)
which is equal to A2 - B2 (you might recall the phrase difference of squares). The squaring eliminates the square roots from the numerator.
As a result, our expression above becomes
x+h- h
x
?
x x
+ +
h h
+ +
x
x
=
x
+
h
-
x
h( x + h + x)
=
h( x
h +h
+
x)
=
h h
x
+
1 h
+
x
=
x
+
1 h
+
. x
Thus, we have shown that
x+h- h
x=
1 .
x+h+ x
This is as simplified as we can make it, and it has the advantage over the original expression in that it has no h multiplier in the denominator (which will be a consideration when you see this sort of thing again in Calculus).
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