Question #aa218

Question

267 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-05T19:24:50

#3x^2+3xt+t^2#, #t ne 0#

Given #f(x)=x^3+1# we want to find #(f(x+t)-f(x))/t#.

To keep it cleaner, first find #f(x+t)#:

#f(x+t) =(x+t)^3+1#

#=x^3+3x^2t+3xt^2+t^3+1#

Now we substitute into #(f(x+t)-f(x))/t#:

#((x^3+3x^2t+3xt^2+t^3+1)-(x^3+1))/t#

#=(x^3+3x^2t+3xt^2+t^3+1-x^3-1)/t#

#=(3x^2t+3xt^2+t^3)/t#

#=3x^2+3xt+t^2#, #t ne 0#