Question

1. consider the graph of y = f(x) provided in figure 1.3.10.

a. on the graph of y = f(x), sketch and label the following quantities:

• the secant line to y = f(x) on the interval -3, -1 and the secant line to y = f(x) on the interval 0, 2.
• the tangent line to y = f(x) at x = -3 and the tangent line to y = f(x) at x = 0.

b. what is the approximate value of the average rate of change of f on -3, -1? on 0, 2? how are these values related to your work in (a)?
c. what is the approximate value of the instantaneous rate of change of f at x = -3? at x = 0? how are these values related to your work in (a)?

Explanation:

Step1: Recall secant - line concept

The secant line between two points \((x_1,y_1)\) and \((x_2,y_2)\) on \(y = f(x)\) is a straight - line connecting these two points. For the interval \([-3,-1]\), find the points \((-3,f(-3))\) and \((-1,f(-1))\) on the graph of \(y = f(x)\) and draw a straight line between them. Similarly, for the interval \([0,2]\), find the points \((0,f(0))\) and \((2,f(2))\) and draw a straight line between them.

Step2: Recall tangent - line concept

The tangent line to \(y = f(x)\) at \(x=a\) is the line that touches the curve \(y = f(x)\) at the point \((a,f(a))\) and has the slope equal to \(f^{\prime}(a)\). For \(x=-3\), draw a line that just touches the curve at the point \((-3,f(-3))\). For \(x = 0\), draw a line that just touches the curve at the point \((0,f(0))\).

Step3: Calculate average rate of change

The average rate of change of \(y = f(x)\) on the interval \([x_1,x_2]\) is given by \(\frac{f(x_2)-f(x_1)}{x_2 - x_1}\). For the interval \([-3,-1]\), the average rate of change is \(\frac{f(-1)-f(-3)}{-1-(-3)}=\frac{f(-1)-f(-3)}{2}\). For the interval \([0,2]\), the average rate of change is \(\frac{f(2)-f(0)}{2 - 0}=\frac{f(2)-f(0)}{2}\). The slope of the secant line on an interval \([x_1,x_2]\) is equal to the average rate of change of the function on that interval.

Step4: Estimate instantaneous rate of change

The instantaneous rate of change of \(y = f(x)\) at \(x=a\) is \(f^{\prime}(a)\), which is the slope of the tangent line at \(x = a\). To estimate \(f^{\prime}(-3)\), we look at the slope of the tangent line at \(x=-3\). To estimate \(f^{\prime}(0)\), we look at the slope of the tangent line at \(x = 0\).

Answer:

a. Sketch the secant and tangent lines as described above.
b. The average rate of change on \([-3,-1]\) is \(\frac{f(-1)-f(-3)}{2}\), and on \([0,2]\) is \(\frac{f(2)-f(0)}{2}\). These values are equal to the slopes of the respective secant lines.
c. The instantaneous rate of change at \(x=-3\) is \(f^{\prime}(-3)\) (slope of the tangent line at \(x=-3\)) and at \(x = 0\) is \(f^{\prime}(0)\) (slope of the tangent line at \(x = 0\)). These values are the slopes of the respective tangent lines.