If # g(x)=x^2+3x+1 # then show that #g(x+1)-g(x) = 2x+4#?

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Answer 1 · 2017-03-13T16:43:43

#(i) : g(x+1)=x^2+5x+5.#

#(ii) :# For, Verification, refer to The Explanation Section.

#g(x)=x^2+3x+1.#

#rArr g(x+1)=(x+1)^2+3(x+1)+1.#

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

#:. g(x+1)=x^2+5x+5.#

Next, #g(x+1)-g(x)=x^2+5x+5-(x^2+3x+1)#

#rArr g(x+1)-g(x)=2x+4.# Hence, the Verification.

Answer 2 · 2017-03-13T16:44:36

We have;

# g(x)=x^2+3x+1 #

And so:

# g(x+1) = (x+1)^2+3(x+1)+1 #
# " " = (x^2+2x+1)+(3x+3)+1 #
# " " = x^2+5x+5 #

Therefore:

# g(x+1) -g(x) = (x^2+5x+2) - (x^2+3x+1) #
# " " = x^2+5x+5 - x^2-3x-1 #
# " " = 2x+4 \ \ \ \ \ QED #