Difference quotient at a point

Differentiation, derivatives and Taylor approximations: Tangent line

Difference quotient at a point

Using the difference quotient, we can approximate the change at one point of a graph. All we have to do is fix the left point of the interval $[a,b]$ and make the length ${\vartriangle}t$ of the interval smaller and smaller.
For a smooth function $f$ the difference quotient will then be close to a certain number, say $m$ . We call this number $m$ the slope of the graph at the point $\bigl(a,f(a)\bigr)$ . As we will see later, this number $m$ is the slope of the tangent to the graph of $f$ at the point $\bigl(a,f(a)\bigr)$ . Smoothness of the function at a point $\bigl(a,f(a)\bigr)$ is worded in mathematics as differentiability at a point, i.e. the existence of the limit $\displaystyle \lim_{{\vartriangle}\to 0}\dfrac{f(a+{\vartriangle}t)-f(a)}{{\vartriangle}t}$ .

For the quadratic polynomial function $f(t)=3t^2-2t-1$ we calculate the difference quotient at the intervals $[1,1.1]$ , $[1,1.01]$ , $[1,1.001]$ and $[1,1.0001]$
$\begin{aligned} {\vartriangle}t &= 0.1 & \frac{{\vartriangle}f}{{\vartriangle}t} &= \frac{f(1.1)-f(1)}{0.1}=\frac{0.433-0}{0.1}=4.3\\ \\{\vartriangle}t &= 0.01 & \frac{{\vartriangle}f}{{\vartriangle}t} &= \frac{f(1.01)-f(1)}{0.01}=\frac{0.0403-0}{0.01}=4.03\\ \\ {\vartriangle}t &= 0.001 & \frac{{\vartriangle}f}{{\vartriangle}t} &= \frac{f(1.001)-f(1)}{0.001}=\frac{0.004003-0}{0.001}=4.003\\ \\ {\vartriangle}t &= 0.0001 & \frac{{\vartriangle}f}{{\vartriangle}t} &= \frac{f(1.0001)-f(1)}{0.0001}=\frac{0.00040003-0}{0.0001}=4.0003 \end{aligned}$
You now suspect that $\displaystyle \frac{{\vartriangle}f}{{\vartriangle}t}\to 4$ on the interval $1,1+{\vartriangle}t$ as ${\vartriangle}t\to 0$ . We will verify this later.

You can also explore this in the interactive example below for the quadratic function.

The interactive diagram on the right-hand side shows the graph of the function $f(t)= t^2$ with two points on the graph. The bottom point $A=\bigl(t_A,f(t_A)\bigr)$ on the graph is free to move by dragging the magenta coloured point on the horizontal axis and the top point $B=\bigl(t_B,f(t_B)\bigr)$ on the graph is obtained by $t_B=t_A+{\vartriangle}t$ for a horizontal change ${\vartriangle}t$ . The horizontal change ${\vartriangle}t$ is therefore: ${\vartriangle}t =t_B-t_A$ The corresponding vertical change is: $\begin{aligned} {\vartriangle}f&=f(t_B) -f(t_A)\\[0.2cm] &= f(t_A+{\vartriangle}t)-f(t_A)\end{aligned}$ The average rate of change $\frac{{\vartriangle}f}{{\vartriangle}t}$ over the interval $[t_A,t_B]$ in this example does depend on the choice of the point $A$ and the interval (or if you prefer the horizontal change ${\vartriangle}t$ ); just move the slider or the point $A$ to observe this. As you go a smaller horizontal change ${\vartriangle}t$ with a fixed choice of the point $A$ the average rate of change $\frac{{\vartriangle}f}{{\vartriangle}t}$ goes towards a constant.

GeoGebra