How do you compute the rate of change of the function f(x) = 1+1/(x+1) at the point x = a?

Aug 3, 2017

Using the definition, see below.

Explanation:

The rate of change at $x = a$ can be found using

${\lim}_{h \rightarrow 0} \frac{f \left(a + h\right) - f \left(a\right)}{h}$

or

${\lim}_{x \rightarrow a} \frac{f \left(x\right) - f \left(a\right)}{x - a}$.

Using the second form, we have

${\lim}_{x \rightarrow a} \frac{f \left(x\right) - f \left(a\right)}{x - a} = {\lim}_{x \rightarrow a} \frac{\left(1 + \frac{1}{x + 1}\right) - \left(1 + \frac{1}{a + 1}\right)}{x - a}$

$= {\lim}_{x \rightarrow a} \frac{\frac{1}{x + 1} - \frac{1}{a + 1}}{x - a}$

$= {\lim}_{x \rightarrow a} \frac{\frac{\left(a + 1\right) - \left(x + 1\right)}{\left(x + 1\right) \left(a + 1\right)}}{\frac{x - a}{1}}$

$= {\lim}_{x \rightarrow a} \frac{a - x}{\left(x + 1\right) \left(a + 1\right)} \cdot \frac{1}{x - a}$

 = lim_(xrarra)(-1(cancel(x-a)))/((x+1)(a+1)) * 1/(1(cancel(x-a))

$= \frac{- 1}{\left(a + 1\right) \left(a + 1\right)} = \frac{- 1}{a + 1} ^ 2$