If #f(x)=1/(x+1)# how do you evaluate #(f(x+h)-f(x))/h#?

Question

4250 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-02T12:08:23

#(f(x+h)-f(x))/h=(-1)/((x+h+1)(x+1))#

As #f(x)=1/(x+1)#, f(x+h)=1/(x+h+1)# and

#(f(x+h)-f(x))/h#

= #(1/(x+h+1)-1/(x+1))/h#

= #((x+1-x-h-1)/((x+h+1)(x+1)))/h#

= #((-h)/((x+h+1)(x+1)))/h#

= #(-h)/((x+h+1)(x+1))xx1/h#

= #(-1)/((x+h+1)(x+1))#

Answer 2 · 2016-08-02T12:17:37

#-1/((x+1)(x+h+1))#

#f(x)=1/(x+1) rArr f(x+h)=1/(x+h+1)#

#rArr f(x+h)-f(x)=1/(x+h+1)-1/(x+1)=(x+1-x-h-1)/((x+1)(x+h+1))#.

#=- h/((x+1)(x+h+1))#

#rArr (f(x+h)-f(x))/h=-cancelh/(cancelh(x+1)(x+h+1))#

#=- 1/((x+1)(x+h+1))#.