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

1 Answer
Jim H
Aug 3, 2017

Using the definition, see below.

Explanation:

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

#lim_(hrarr0)(f(a+h) - f(a))/h#

or

#lim_(xrarra)(f(x) - f(a))/(x-a)#.

Using the second form, we have

#lim_(xrarra)(f(x) - f(a))/(x-a) = lim_(xrarra)((1+1/(x+1)) - (1+1/(a+1)))/(x-a)#

# = lim_(xrarra)(1/(x+1) - 1/(a+1))/(x-a)#

# = lim_(xrarra)(((a+1)-(x+1))/((x+1)(a+1)))/((x-a)/1)#

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

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

# = (-1)/((a+1)(a+1)) = (-1)/(a+1)^2#

Related questions
See all questions in Instantaneous Rate of Change at a Point
Impact of this question
1862 views around the world
You can reuse this answer
Creative Commons License