Given #g(x)=5/(x−3)#, evaluate and simplify: #(g(5+h)−g(5))/h =#?

2 Answers
Tazwar Sikder
Mar 16, 2018

#- frac(5)(2h + 4)#

Explanation:

We have: #g(x) = frac(5)(x - 3)#

In order to evaluate #frac(g(5 + h) - g(5))(h)#, we simply substitute #5 + h# and #5# in place of #x# in #g(x)#:

#Rightarrow frac(g(5 + h) - g(5))(h) = frac(frac(5)((5 + h) - 3) - frac(5)((5) - 3))(h)#

#Rightarrow frac(g(5 + h) - g(5))(h) = frac(frac(5)(h + 2) - frac(5)(2))(h)#

#Rightarrow frac(g(5 + h) - g(5))(h) = frac(frac(5 cdot 2 - 5 cdot (h + 2))(2 cdot (h + 2)))(h)#

#Rightarrow frac(g(5 + h) - g(5))(h) = frac(frac(10 - 5h - 10)(2 cdot (h + 2)))(h)#

#Rightarrow frac(g(5 + h) - g(5))(h) = frac(- 5h)(2 cdot (h + 2)) cdot frac(1)(h)#

#therefore frac(g(5 + h) - g(5))(h) = - frac(5)(2h + 4)#

Jim G.
Mar 16, 2018

#-5/(2(h+2))#

Explanation:

#"evaluating each term separately"#

#g(5+h)=5/(5+h-3)=5/(2+h)#

#f(5)=5/(5-3)=5/2#

#rArr(g(5+h)-g(5))#

#=5/(2+h)-5/2#

#=(10-5(2+h))/(2(2+h))=(-5h)/(2(2+h))#

#rArr(g(5+h)-g(5))/h#

#=(-5cancel(h))/(cancel(h)2(2+h))=-5/(2(2+h))#

Related questions
Impact of this question
6561 views around the world
You can reuse this answer
Creative Commons License