Activity 7.
Consider the function \(f\) whose formula is \(\displaystyle f(x) = 3 - 2x\text{.}\)
(a)
What familiar type of function is \(f\text{?}\) What can you say about the slope of \(f\) at every value of \(x\text{?}\)
Hint.
If \(f(x) = 3x^2 + 2x - 4\text{,}\) we say “\(f\) is quadratic.” If \(f(x) = 5 e^{2x-1}\text{,}\) we say “\(f\) is exponential.” What do we say about \(f(x) = 3-2x\text{?}\)
Answer.
\(f\) is linear.
Solution.
Because \(f(x) = 3 - 2x\) is of the form \(f(x) = mx + b\text{,}\) we call \(f\) a linear function.
(b)
Compute the average rate of change of \(f\) on the intervals \([1,4]\text{,}\) \([3,7]\text{,}\) and \([5,5+h]\text{;}\) simplify each result as much as possible. What do you notice about these quantities?
Hint.
Remember that to compute the average rate of change of \(f\) on \([a,b]\text{,}\) we calculate \(\frac{f(b)-f(a)}{b-a}\text{.}\)
Answer.
The average rate of change on \([1,4]\text{,}\) \([3,7]\text{,}\) and \([5,5+h]\) is \(-2\text{.}\)
Solution.
The average rate of change on \([1,4]\) is \(\frac{f(4)-f(1)}{4-1} = \frac{-5 - 1}{3} = -2\text{.}\) Similar calculations show the average rate of change on \([3,7]\) is also \(-2\text{.}\) On \([5,5+h]\text{,}\) observe that
\begin{align*}
\frac{f(5+h)-f(5)}{h} \amp = \frac{3-2(5+h) - (3-10)}{h}\\
\amp = \frac{3 - 10 - 2h + 7}{h}\\
\amp = \frac{-2h}{h}\\
\amp = -2\text{.}
\end{align*}
(c)
Use the limit definition of the derivative to compute the exact instantaneous rate of change of \(f\) with respect to \(x\) at the value \(a = 1\text{.}\) That is, compute \(f'(1)\) using the limit definition. Show your work. Is your result surprising?
Hint.
Observe that \(f(1+h) = 3 - 2(1+h) = 3 - 2 - 2h = 1 - 2h\text{.}\)
Answer.
\(f'(1)=-2\text{.}\)
Solution.
Using the limit definition of the derivative, we find that
\begin{align*}
f'(1) = \amp \lim_{h \to 0} \frac{f(1+h) - f(1)}{h}\\
= \amp \lim_{h \to 0} \frac{(3 - 2(1+h)) - (3-2)}{h}\\
= \amp \lim_{h \to 0} \frac{3 - 2 - 2h - 1}{h}\\
= \amp \lim_{h \to 0} \frac{-2h}{h}\\
= \amp \lim_{h \to 0} -2\\
= \amp -2\text{.}
\end{align*}
(d)
Without doing any additional computations, what are the values of \(f'(2)\text{,}\) \(f'(\pi)\text{,}\) and \(f'(-\sqrt{2})\text{?}\) Why?
Hint.
Think about the how the graph of \(f\) appears. What is the same at every point?
Answer.
\(f'(2)=-2\text{,}\) \(f'(\pi)=-2\text{,}\) and \(f'(-\sqrt{2})=-2\text{,}\) since the slope of a linear function is the same at every point.
Solution.
\(f'(2)=-2\text{,}\) \(f'(\pi)=-2\text{,}\) and \(f'(-\sqrt{2})=-2\text{,}\) since the slope of a linear function is the same at every point.
