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

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Answer 1 · 2017-03-13T16:43:43

$(i) : g (x + 1) = x^{2} + 5 x + 5 .$ $(i) : g(x+1)=x^2+5x+5.$

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

$g (x) = x^{2} + 3 x + 1 .$ $g(x)=x^2+3x+1.$

$\Rightarrow g (x + 1) = {(x + 1)}^{2} + 3 (x + 1) + 1 .$ $rArr g(x+1)=(x+1)^2+3(x+1)+1.$

$= x^{2} + 2 x + 1 + 3 x + 3 + 1$ $=x^2+2x+1+3x+3+1$

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

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:

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

If g(x)=x^2+3x+1 g(x)=x2+3x+1 g(x)=x^2+3x+1 then show that g(x+1)-g(x) = 2x+4g(x+1)−g(x)=2x+4g(x+1)-g(x) = 2x+4?