Chapter 5.1, Problem 79STP

To determine

To show:

The line through a general form of an equation

Explanation of Solution

Given:

The function $f\left(x\right)=y=4$ is a linear function then $f\text{'}\left(x\right)=0$

Concept used:

The point on the line through $P\left(x,y\right)$ and the line of the slope is $m$

The line of the slope of the equation $y=mx+b\mathrm{..........}\left(1\right)$

Calculation:

The function

  $f\left(x\right)=y=4$

The line of the slope of the equation $y=mx+b\mathrm{..........}\left(1\right)$

Where b is any constant

  $\begin{array}{l}f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f\left(x+h\right)-f\left(x\right)}{h}\\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{\left(m\left(x+h\right)+b\right)-\left(mx+b\right)}{h}\\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{\left(mx+mh\right)+b-\left(mx+b\right)}{h}\\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{\left(mx+mh\right)+b-mx-b}{h}\\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{\left(mh\right)}{h}\\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{m\left(h\right)}{h}\\ f\text{'}\left(x\right)=\underset{h\to 0}{\lim }m\\ f\text{'}\left(x\right)=m\end{array}$

The tangent line to $f$ at any point $x=c$

If the slope $m$ of a line and a point $x=c$ on the line then the equation of the line $y=mx+b$

Putting the value of $x=c$

  $\begin{array}{l}0=mc+b\\ b=-mc\end{array}$

Putting the value of b in equation (1)

  $\begin{array}{l}y=mx-mc\\ y=m\left(x-c\right)\end{array}$

  $\begin{array}{l}f\left(x\right)=y=4\\ y=4\end{array}$

Hence, the function $f\left(x\right)=y=4$ is a linear function then $f\text{'}\left(x\right)=0$