Let #f(x) = 9x^2-x#. What is #f(x+1)-f(x)#?

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Answer 1 · 2018-01-26T13:52:08

See a solution process below:

First, let's find #f(x + 1)# by substituting #(color(red)(x + 1))# for every occurrence of #color(blue)(x)# in #f(x)#:

#f(color(blue)(x)) = 9color(blue)(x)^2 - color(blue)(x)# becomes:

#f(color(red)(x + 1)) = 9(color(red)(x + 1))^2 - (color(red)(x + 1))#

#f(color(red)(x + 1)) = 9(color(red)(x)^2 + color(red)(2x) + color(red)(1))^2 - color(red)(x) - color(red)(1)#

#f(x + 1) = 9x^2 + 18x + 9 - x - 1#

#f(x + 1) = 9x^2 + 18x - x + 9 - 1#

#f(x + 1) = 9x^2 + 18x - 1x + 9 - 1#

#f(x + 1) = 9x^2 + 17x + 8#

Now, we can determine #f(x + 1) - f(x)#:

#f(x + 1) - f(x) => (9x^2 + 17x + 8) - (9x^2 - x) =>#

#9x^2 + 17x + 8 - 9x^2 + x =>#

#9x^2 - 9x^2 + 17x + x + 8 =>#

#0 + 18x + 8 =>#

#18x + 8#

Or

#2(9x + 4)#