Question

normal line to the curve at the point given.

1. $f(x)=\frac{1}{x^{3}}$ at $(1,1)$
2. $f(x)=4+\sqrt{x}$ at $(4,6)$

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 the given point

Substitute \(x = 1\) into \(f^\prime(x)\).
\(f^\prime(1)=-\frac{3}{1^{4}}=-3\).

Step3: Find the slope of the normal line

The slope of the normal line \(m_{normal}\) is the negative reciprocal of the slope of the tangent line.
Since the slope of the tangent line \(m_{tangent}=f^\prime(1)=-3\), then \(m_{normal}=\frac{1}{3}\).

Step4: Use the point - slope form to find the equation of the normal line

The point - slope form is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})=(1,1)\) and \(m = \frac{1}{3}\).
\(y - 1=\frac{1}{3}(x - 1)\).
Expand it: \(y-1=\frac{1}{3}x-\frac{1}{3}\).
\(y=\frac{1}{3}x-\frac{1}{3}+1=\frac{1}{3}x+\frac{2}{3}\).

Answer:

\(y=\frac{1}{3}x+\frac{2}{3}\)