the curve below is called a witch of maria agnesi. find an equation of the tangent line to this curve at th…

Explanation:

Step1: Find the derivative of $y$

Use the quotient - rule. If $y=\frac{u}{v}$ where $u = 1$ and $v=1 + x^{2}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Since $u^\prime = 0$ and $v^\prime=2x$, we have $y^\prime=\frac{0\times(1 + x^{2})-1\times2x}{(1 + x^{2})^{2}}=-\frac{2x}{(1 + x^{2})^{2}}$.

Step2: Let the point of tangency be $(a,b)$.

First, $b=\frac{1}{1 + a^{2}}$. The slope of the tangent line at $x = a$ is $m=-\frac{2a}{(1 + a^{2})^{2}}$.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$. Here $(x_1,y_1)=(a,\frac{1}{1 + a^{2}})$ and $m = -\frac{2a}{(1 + a^{2})^{2}}$. So $y-\frac{1}{1 + a^{2}}=-\frac{2a}{(1 + a^{2})^{2}}(x - a)$. Simplify: [ \begin{align*} y&=\frac{1}{1 + a^{2}}-\frac{2a}{(1 + a^{2})^{2}}(x - a)\ &=\frac{1}{1 + a^{2}}-\frac{2ax}{(1 + a^{2})^{2}}+\frac{2a^{2}}{(1 + a^{2})^{2}}\ &=-\frac{2ax}{(1 + a^{2})^{2}}+\frac{1 + 2a^{2}}{(1 + a^{2})^{2}} \end{align*} ] If we assume the point of tangency is not given and we want the general form of the tangent - line equation in terms of a general $x$ - value $x_0$: The slope of the tangent line at $x = x_0$ is $m=-\frac{2x_0}{(1 + x_0^{2})^{2}}$, and the point on the curve is $(x_0,\frac{1}{1 + x_0^{2}})$. The point - slope form $y-\frac{1}{1 + x_0^{2}}=-\frac{2x_0}{(1 + x_0^{2})^{2}}(x - x_0)$. [ \begin{align*} y&=\frac{1}{1 + x_0^{2}}-\frac{2x_0}{(1 + x_0^{2})^{2}}(x - x_0)\ &=\frac{1}{1 + x_0^{2}}-\frac{2x_0x}{(1 + x_0^{2})^{2}}+\frac{2x_0^{2}}{(1 + x_0^{2})^{2}}\ &=-\frac{2x_0x}{(1 + x_0^{2})^{2}}+\frac{1 + 2x_0^{2}}{(1 + x_0^{2})^{2}} \end{align*} ] Let's assume we want the tangent line at the point $x = 1$. When $x = 1$, $y=\frac{1}{1+1}=\frac{1}{2}$, and $y^\prime=-\frac{2\times1}{(1 + 1)^{2}}=-\frac{1}{2}$. Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(1,\frac{1}{2})$ and $m=-\frac{1}{2}$, we have $y-\frac{1}{2}=-\frac{1}{2}(x - 1)$. [ \begin{align*} y-\frac{1}{2}&=-\frac{1}{2}x+\frac{1}{2}\ y&=-\frac{1}{2}x + 1 \end{align*} ]

Answer:

If the point of tangency is $(1,\frac{1}{2})$, the equation of the tangent line is $y=-\frac{1}{2}x + 1$. In general, for a point of tangency $(x_0,\frac{1}{1 + x_0^{2}})$ the equation of the tangent line is $y=-\frac{2x_0x}{(1 + x_0^{2})^{2}}+\frac{1 + 2x_0^{2}}{(1 + x_0^{2})^{2}}$.