# Math Insight

### Approximating a nonlinear function by a linear function

Math 1241, Fall 2020
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Due date: Oct. 9, 2020, 11:59 p.m.
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Total points: 3
1. Let $g(x) = x^3$. Find the average rate of change in $g(x)$ between:
1. $x=3$ and $x=4$.

2. $x=3$ and $x=3.1$.
3. $x=3$ and $x=3.01$.
4. $x=2$ and $x=3$.
5. $x=2.9$ and $x=3$.
6. $x=2.99$ and $x=3$.
7. $x=a$ and $x=b$
8. $x=a$ and $x=a+h$

2. Feedback from applet
number of secants:
number of tangents:
secants:
tangents:
Let $f(x) = - x^{2} + 10 x$, which is graphed to the right.
1. Sketch a secant line from $x=2$ to $x=4$. What is its slope?
2. Sketch a tangent line at $x=2$. Estimate its slope by calculating the slope of secant lines from $x=2$ to $x=2+h$ for the following values of $h$:
• $h=0.1$:
• $h= 0.01$:
• $h= 0.001$:
• $h=-0.1$:
• $h= -0.01$:
• $h= -0.001$:
(Round above answers to nearest 1000th or less.)

As $h$ approach zero, it appears the slope approaches what value?

Based on this observation, estimate the the slope of the tangent line at $x=2$:
$f'(2) =$
.

3. Sketch a tangent line at $x=4$. Estimate its slope by calculating the slope of secant lines from $x=4$ to $x=4+h$ for the following values of $h$:
• $h=0.1$:
• $h= 0.01$:
• $h= 0.001$:
• $h=-0.1$:
• $h= -0.01$:
• $h= -0.001$:
(Round above answers to nearest 1000th or less.)

The slope of the tangent line appears to be $\displaystyle\diff{f}{x}\bigg|_{x=4} =$
.

3. A squirrel is running along a branch avoiding an oncoming owl. Her position $x$ along the branch after $t$ seconds is $g(t)=4 t^{5} - 52 t^{4} + 236 t^{3} - 424 t^{2} + 240 t$, where $g(t)$ is given in cm, relative to her position at time $t=0$ seconds. The graph of $x=g(t)$ is shown below.
Feedback from applet
number of secants:
number of tangents:
secants:
tangents:
1. After five seconds, her position along the branch is $g(5)=100$ cm. What is the average rate of change of her position (i.e., her average velocity) during the five seconds from $t=0$ to $t=5$?
cm/second.

Sketch a secant line on the above graph that shows her position as a function of time if she had moved steadily at that velocity during those five seconds. What is the equation of that line?
$x=$

2. Her position at $t=1$ seconds is $g(1)=4$ cm; at $t=2$ seconds, it is $g(2)=-32$ cm. What is her average velocity during the period between $1 \le t \le 2$?
cm/second.

Sketch a secant line on the above graph, which would show her position if she had moved steadily at that velocity starting at $t=1$. What is the equation of that line?
$x=$

3. Approximate the instantaneous rate of her position (her velocity) at $t=3$ seconds. To do so, calculate her average rate of change from $t=3$ to $t=3+\Delta t$ for $\Delta t=0.01$ and $0.001$.

Estimated velocity from $\Delta t=0.01$:
cm/second.
Estimated velocity from $\Delta t=0.001$:
cm/second.
(Round the above answers to nearest 100th of a cm/second or less.)

As $\Delta t$ decreases, it appears the velocity is approaching what integer value?
cm/second

4. Assuming she continued moving steadily at this velocity starting at $t=3$ seconds, write down an equation for her position as a function of time.
$x=$

The graph of this equation will be a tangent line to the graph at $t=3$ seconds. Sketch the tangent line on the above graph.

4. The slope of a graph at a point is the same thing as the slope of the tangent line at that point. Estimate the slope of the function $f(x)=x^{2}$ at $x=1$ by calculating the slope of the secant line between $x=1$ and $x=1+\Delta x$ for the following values of $\Delta x$:
• $\Delta x = 0.1$:
• $\Delta x = 0.01$:
• $\Delta x = 0.001$:
• $\Delta x = -0.1$:
• $\Delta x = -0.01$:
• $\Delta x = -0.001$:
(Round answer to nearest 1000th or less.)

Since for both positive $\Delta x$ and negative $\Delta x$, the slope of the secant is approaching the same value as $\Delta x$ approaches zero, the slope of the function at $x=1$ must be
.