Question 1005663
The slope of the tangent line is equal to the value of the derivative at that point.
So first find the derivative using the quotient rule.
{{{df/dx=((x^2+x+1)(3x^2)-(2x+1)(x^3+1))/(x^2+x+1)^2}}}
I won't bother to simplify since we're doing an evaluation,
So when, {{{x=1}}},
{{{m=df/dx=((1^2+1+1)(3(1)^2)-(2(1)+1)(1^3+1))/(1^2+1+1)^2}}}
{{{m=(3(3)-3(2))/(3)^2}}}
{{{m=(9-6)/9}}}
{{{m=1/3}}}
So then find the value of the function at {{{x=1}}}
{{{y=(1^3+1)/(1^2+1+1)=2/3}}}
Use the point-slope form of a line,
{{{y-2/3=(1/3)(x-1)}}}
{{{y-2/3=(1/3)x-1/3}}}
{{{y=(1/3)x+1/3}}}
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