Question

Evaluate the difference quotient for the given function.
f(x) = 1/x, [f(x) - f(a)]/(x-a)

I don't understand how you start this off at all. I know that f(a) = 1/a but I don't understand how you get that, either.

Answer 1

\[\Large \frac{f(x) - f(a)}{x-a}\]



\[\Large \frac{\frac{1}{x} - \frac{1}{a}}{x-a}\]



\[\Large \frac{ax(\frac{1}{x}) - ax(\frac{1}{a})}{ax(x)-ax(a)}\]



\[\Large \frac{a-x}{ax^2-a^2x}\]



\[\Large \frac{a-x}{-ax(a-x)}\]



\[\Large \frac{\cancel{a-x}}{-ax\cancel{(x-a)}}\]



\[\Large -\frac{1}{ax}\]

So \[\Large \frac{f(x) - f(a)}{x-a}=-\frac{1}{ax}\] when \[\Large f(x)=\frac{1}{x}\]

Answer 2

\[\Large f(a) = \frac{1}{a}\] because all that we did was replace 'x' with 'a'

Answer 3

how do you come across multiplying the numerator and denominator by ax/how do you determine that's the common denominator?

Answer 4

the LCD of 1/x and 1/a is ax, multiplying EVERY term by this clears out the inner fractions

Answer 5

the LCD is the LCM of the denominators


to find the LCM of a and x, just multiply them and then divide by the GCF. Since the GCF is 1, we get (ax)/1 = ax