Q. 64 - Calculus

Question

Use the definition of the derivative to find the equations of the lines described in Exercises 59-64.

The line that is perpendicular to the tangent line to $f\left(x\right)=x^4+1$ at $x=2$ and also passes through the point $(-1,8)$

Step-by-Step Solution

Answer

The equation is $y=-\frac{1}{32}x+\frac{255}{32}$

Step 1. Given information

Given the function $f\left(x\right)=x^4+1$ and the point $(-1,8)$

Step 2: Calculate f ' ( 2 ) using definition of derivative

Calculating, we get

$$\begin{gathered} f^{\text{'}}\left(2\right)=\underset{x\to 2}{\lim }\frac{f\left(x\right)-f\left(2\right)}{x-2} \\ {f}^{\text{'}}\left(2\right)=\underset{x\to 2}{\lim }\frac{x^4+1-\left(2^4+1\right)}{x-2} \\ f\text{'}\left(2\right)=\underset{x\to 2}{\lim }\frac{x^4-2^4}{x-2} \\ f\text{'}\left(2\right)=\underset{x\to 2}{\lim }\frac{(x-2)(x+2)\left(x^2+2^2\right)}{(x-2)} \\ f\text{'}\left(2\right)=\underset{x\to 2}{\lim }(x+2)\left(x^2+2^2\right) \\ f\text{'}\left(2\right)=(2+2)\left(2^2+2^2\right) \\ f\text{'}\left(2\right)=32 \end{gathered}$$

Step 3: Use point slope form and calculate the form

Calculating, we get

Line is perpendicular to tangent line so $m=-\frac{1}{32}$

$$\begin{gathered} y-8=-\frac{1}{32}(x+1) \\ y=-\frac{1}{32}x-\frac{1}{32}+8 \\ y=-\frac{1}{32}x+\frac{255}{32} \end{gathered}$$