Question

mal line to the curve at the point given.

1. $f(x)=\frac{1}{x^{3}}$ at $(1,1)$

Explanation:

Step1: Find the derivative of the function

The function \(f(x)=\frac{1}{x^{3}}=x^{-3}\). Using the power rule \((x^{n})^\prime = nx^{n - 1}\), we have \(f^\prime(x)=-3x^{-4}=-\frac{3}{x^{4}}\).

Step2: Evaluate the derivative at \(x = 1\)

Substitute \(x = 1\) into \(f^\prime(x)\). \(f^\prime(1)=-\frac{3}{1^{4}}=-3\). The slope of the tangent line at \((1,1)\) is \(m_{tangent}=-3\).

Step3: Find the slope of the normal line

Since the slope of the normal line \(m_{normal}\) and the slope of the tangent line \(m_{tangent}\) satisfy \(m_{normal}\times m_{tangent}=-1\). Given \(m_{tangent}=-3\), then \(m_{normal}=\frac{1}{3}\).

Step4: Use the point - slope form \(y - y_{1}=m(x - x_{1})\)

Here \(x_{1}=1,y_{1}=1,m = \frac{1}{3}\). So \(y - 1=\frac{1}{3}(x - 1)\). Simplify it: \(y-1=\frac{1}{3}x-\frac{1}{3}\), then \(y=\frac{1}{3}x+\frac{2}{3}\).

Answer:

The equation of the normal line is \(y=\frac{1}{3}x+\frac{2}{3}\)