Question

Use the definition of the derivative to prove that our concept of slope for linear functions matches the slope that is defined by the derivative. In other words, show that if $f\left(x\right)=mx+b$is any linear function, then $f\text{'}\left(x\right)=m$.

Short answer

The slope intercept form a linear function is $mx+b$, where $m$is the slope of the function.

Also by using the derivative definition the derivative of function $mx+b$is $m$same as the slope.

So it is proved that the slope of a linear function is defined by the derivative.

Step-by-step solution

Step 1. Given Information

We have given the following linear function :-

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

We have to prove the slope of this function is defined by its derivative.

Step 2. Slope of the function

We have given the following linear function :-

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

This is the general linear function in slope intercept form.

So that slope of the function is$m$.

Step 3. To find derivative

The given function is :-

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

We know that the derivative of a function $f\left(x\right)$is defined as :-

$f\text{'}\left(x\right)=\underset{h\to 0}{\lim }\frac{f(x+h)-f\left(x\right)}{h}$

Put all the values :-

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

This is same as the slope of the function.

So that we can conclude that slope of a linear function is defined by its derivative.