What is the instantaneous rate of change for $ƒ(x) = 2x + 5$ at $x = -7$? What is the instantaneous rate of change for $ƒ(x) = 2x + 5$ at $x = -7$ ?
Do I just simply substitute $-7$ into the equation to get the answer? Would that mean the solution is $-9$?
 A: 

*The rate of change for a line is the slope, so the rate of change along the entire line is 2.

A: The instantaneous rate of change of a function at a point is the function's derivative at that point.  
The function $f(x) = 2x + 7$ has derivative $f'(x) = 2$.  Since the derivative does not depend on $x$, the instantaneous rate of change at every point of the linear function $f(x) = 2x + 7$ is $2$, the slope of the line $y = 2x + 7$.  In particular, the instantaneous rate of change of change at $x = -7$ is $f'(-7) = 2$.  
In general, the derivative of a function at a point represents the slope of the tangent line to its graph at that point.  For a linear function $f(x) = mx + b$, the derivative $f'(x) = m$ is just the slope of the line $y = mx + b$.
A: Let $h$ be equal to a very small change in $x$. Thus, the slope of a line at $x$ to $x+h$ is
$$\frac{\Delta f}{\Delta x}=\frac{f(x+h)-f(x)}{x+h-x}=\frac{f(x+h)-f(x)}{h}$$
To find the instantaneous rate, we wish to find the limit as $h$ approaches $0$.
$$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$$
For our function:
$$\lim_{h \to 0}\frac{2(x+h)-5-(2x-5)}{h}=\lim_{h \to 0}\frac{2x+2h-5-2x+5}{h}=\lim_{h \to 0}\frac{2h}{h}=\lim_{h \to 0}2=2.$$
Thus, at any $x$ value, including $x=-7$, the instantaneous rate of change is $2$.
A: The line has a constant slope, or gradient, or rate of change equaling its derivative =2. $x$ does not occur anymore in this example to plug in.
