Question

Use Problem 93 to prove that a linear function is its own tangent line at every point. In other words, show that if $f\left(x\right)=mx+b$is any linear function, then the tangent line to$f$at any point $x=c$is given by $y=mx+b$.

Short answer

Ans:

$$\begin{gathered} y-f\left(c\right)=f^{\text{'}}\left(c\right)[x-c] \\ y-(mc+b)=m[x-c] \\ y-mc-b=mx-mc \\ y=mx-mc+mc+b \\ y=mx+b \end{gathered}$$

Step-by-step solution

Step 1. Given information:

Consider the function:

$f\left(x\right)=mx+b$

Here the objective is to show that the equation of the tangent line at any point $x=c$is $y=mx+b$.

Step 2. Solving the equation:

$$\begin{gathered} f^{\text{'}}\left(x\right)=\underset{z\to 4}{\lim }\frac{f\left(z\right)-f\left(x\right)}{z-x} \\ =\underset{z\to x}{\lim }\frac{(mz+b)-(mx+b)}{z-x} \\ =\underset{z\to x}{\lim }\frac{m(z-x)}{z-x} \\ =\underset{z\to x}{\lim }m \\ =m \end{gathered}$$

Hence the slope of the linear function is constant and it is $f\text{'}\left(x\right)=m$

At any point x=c the slope of the function is $f\text{'}\left(c\right)=m$

At x=c, the value of the function is

$f\left(c\right)=mc+b$

Step 3. Finding the equation of the tangent line:

The eq of the tangent line will be,

$$\begin{gathered} y-f\left(c\right)=f^{\text{'}}\left(c\right)[x-c] \\ y-(mc+b)=m[x-c] \\ y-mc-b=mx-mc \\ y=mx-mc+mc+b \\ y=mx+b \end{gathered}$$

Hence the eq of the tangent is same as linear function.