Question

1. -/1 points find an equation of the tangent line to the graph of the given function at the specified point. f(x) = (x^2 - 1)/(x^2 + x + 1), (1, 0) y =

Explanation:

Step1: Use quotient - rule for differentiation

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x^{2}-1$, so $u^\prime=2x$, and $v=x^{2}+x + 1$, so $v^\prime=2x + 1$. Then $f^\prime(x)=\frac{(2x)(x^{2}+x + 1)-(x^{2}-1)(2x + 1)}{(x^{2}+x + 1)^{2}}$.

Step2: Expand the numerator

Expand $(2x)(x^{2}+x + 1)=2x^{3}+2x^{2}+2x$ and $(x^{2}-1)(2x + 1)=2x^{3}+x^{2}-2x - 1$. Then the numerator is $2x^{3}+2x^{2}+2x-(2x^{3}+x^{2}-2x - 1)=2x^{3}+2x^{2}+2x - 2x^{3}-x^{2}+2x + 1=x^{2}+4x + 1$. So $f^\prime(x)=\frac{x^{2}+4x + 1}{(x^{2}+x + 1)^{2}}$.

Step3: Find the slope of the tangent line

Evaluate $f^\prime(x)$ at $x = 1$. Substitute $x = 1$ into $f^\prime(x)$: $f^\prime(1)=\frac{1^{2}+4\times1 + 1}{(1^{2}+1 + 1)^{2}}=\frac{1 + 4+1}{9}=\frac{2}{3}$.

Step4: Use the point - slope form of a line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,0)$ and $m=\frac{2}{3}$. So $y-0=\frac{2}{3}(x - 1)$, which simplifies to $y=\frac{2}{3}x-\frac{2}{3}$.

Answer:

$y=\frac{2}{3}x-\frac{2}{3}$