Question 1086729
<font color="black" face="times" size="3">{{{f(x)=x^2+x+1}}}


{{{f(x+h)=(x+h)^2+(x+h)+1}}} Replace every x with (x+h)


{{{f(x+h)=x^2+2xh+h^2+x+h+1}}} Expand out (x+h)^2


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Now that we know what f(x+h) is equal to, we can find the difference quotient


We'll be making two substitutions. The first substitution will have us replace the "f(x+h)" with the expression we got above


So we go from this


{{{(highlight(f(x+h)) - f(x))/h = (highlight(f(x+h))-f(x))/h}}}


to this


{{{(highlight(f(x+h))-f(x))/h=(highlight(x^2+2xh+h^2+x+h+1)-f(x))/h}}}


Note the red boxes highlighted to show exactly what is replaced and how it looks after the replacement is done


Similarly, we will replace f(x) with x^2+x+1 to go from this


{{{(f(x+h)-highlight(f(x)))/h=(x^2+2xh+h^2+x+h+1-(highlight(f(x))))/h}}}


to this


{{{(f(x+h)-highlight(f(x)))/h=(x^2+2xh+h^2+x+h+1-(highlight(x^2+x+1)))/h}}}


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After the two substitutions, we now have


{{{(f(x+h)-f(x))/h=(x^2+2xh+h^2+x+h+1-(x^2+x+1))/h}}}


Let's simplify the right hand side (RHS)


{{{(f(x+h)-f(x))/h=(x^2+2xh+h^2+x+h+1-(x^2+x+1))/h}}}


{{{(f(x+h)-f(x))/h=(x^2+2xh+h^2+x+h+1-x^2-x-1)/h}}} Distribute the negative through


{{{(f(x+h)-f(x))/h=(2xh+h^2+x+h+1-x-1)/h}}} Note how x^2-x^2 = 0x^2 = 0 so the x^2 terms go away


{{{(f(x+h)-f(x))/h=(2xh+h^2+h+1-1)/h}}} Similarly, x-x = 0x = 0


{{{(f(x+h)-f(x))/h=(2xh+h^2+h)/h}}} And also 1-1 = 0. Each term now has something with an 'h' as a factor


{{{(f(x+h)-f(x))/h=(h(2x+h+1))/h}}} Factor out that common term h


{{{(f(x+h)-f(x))/h=(cross(h)(2x+h+1))/(cross(h))}}} The 'h's cancel on top and bottom (since h/h = 1)


{{{(f(x+h)-f(x))/h=2x+h+1}}}


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The final simplified answer is {{{2x+h+1}}}</font>