part i - communication and understanding
1. answer any two of the following: 4 marks
a. explain the difference between the average rate of change and the instantaneous rate of change.
b. outline the procedure for calculating instantaneous rate of change.
c. outline the procedure for calculating average rate of change.
d. outline the basic rules for evaluation of limits.
2. define any three of the following: 6 marks
a. quadratic formula
b. average rate of change
c. instantaneous rate of change
d. secant line
e. tangent line
f. tangent point
part ii - short answer questions
instructions: read each question carefully and answer to the best of your ability. you do not need to explain your answers for this part.
1. evaluate: 2 marks
a. $lim_{x
ightarrow0}\frac{1}{3}=$
b. $lim_{x
ightarrow0}sqrt{\frac{x^{2}+16}{2(x + 2)}}=$
2. give the value of each statement. if the value does not exist, write \dne\. 8 marks
a. $lim_{x
ightarrow3}f(x)=$
b. $f(3)=$
c. $lim_{x
ightarrow1}f(x)=$
d. $f(1)=$
e. $lim_{x
ightarrow2}f(x)=$
f. $lim_{x
ightarrow - 2^{+}}f(x)=$
g. $lim_{x
ightarrow - 2^{-}}f(x)=$
h. $f(-2)=$

Question

Accepted Answer

### 1. a.
# Brief Explanations:
The average rate of change of a function $y = f(x)$ over an interval $[a,b]$ is the total change in the function's value divided by the change in the input value. It gives an overall sense of how the function changes on that interval. Mathematically, it is $\frac{f(b)-f(a)}{b - a}$. The instantaneous rate of change at a point $x = c$ is the limit of the average - rate of change as the interval around $c$ gets smaller and smaller. It represents the rate of change of the function at that exact moment and is equivalent to the slope of the tangent line to the graph of the function at $x = c$.
# Answer:
The average rate of change of a function $y=f(x)$ over $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$, giving an overall change rate on the interval. The instantaneous rate of change at $x = c$ is the limit of average - rate of change as the interval around $c$ shrinks, equal to the slope of the tangent line at $x = c$.

### 1. b.
# Brief Explanations:
To calculate the instantaneous rate of change of a function $y = f(x)$ at a point $x = a$, first find the difference quotient $\frac{f(a + h)-f(a)}{h}$, where $h$ represents a small change in the input. Then, find the limit of this difference quotient as $h$ approaches 0, i.e., $\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. This limit value is the instantaneous rate of change of the function at $x = a$.
# Answer:
1. Find the difference quotient $\frac{f(a + h)-f(a)}{h}$.
2. Calculate $\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. The result is the instantaneous rate of change at $x = a$.

### 1. c.
# Brief Explanations:
For a function $y = f(x)$, to find the average rate of change over the interval $[x_1,x_2]$, identify the function values $f(x_1)$ and $f(x_2)$ at the endpoints of the interval. Then use the formula $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$. This formula gives the slope of the secant line connecting the points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ on the graph of the function.
# Answer:
1. Identify $f(x_1)$ and $f(x_2)$ for the interval $[x_1,x_2]$.
2. Use the formula $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$ to get the average rate of change.

### 1. d.
# Brief Explanations:
Some basic rules for evaluating limits:
1. **Constant Rule**: $\lim_{x
ightarrow a}c=c$, where $c$ is a constant.
2. **Identity Rule**: $\lim_{x
ightarrow a}x=a$.
3. **Sum - Rule**: $\lim_{x
ightarrow a}(f(x)+g(x))=\lim_{x
ightarrow a}f(x)+\lim_{x
ightarrow a}g(x)$.
4. **Difference Rule**: $\lim_{x
ightarrow a}(f(x)-g(x))=\lim_{x
ightarrow a}f(x)-\lim_{x
ightarrow a}g(x)$.
5. **Product Rule**: $\lim_{x
ightarrow a}(f(x)\cdot g(x))=\lim_{x
ightarrow a}f(x)\cdot\lim_{x
ightarrow a}g(x)$.
6. **Quotient Rule**: $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x
ightarrow a}f(x)}{\lim_{x
ightarrow a}g(x)}$, provided $\lim_{x
ightarrow a}g(x)
eq0$.
7. **Power Rule**: $\lim_{x
ightarrow a}(f(x))^n = (\lim_{x
ightarrow a}f(x))^n$ for a positive integer $n$.
8. **Root Rule**: $\lim_{x
ightarrow a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x
ightarrow a}f(x)}$ for $n$ a positive integer, and when $\lim_{x
ightarrow a}f(x)\geq0$ if $n$ is even.
# Answer:
Constant Rule: $\lim_{x
ightarrow a}c = c$; Identity Rule: $\lim_{x
ightarrow a}x=a$; Sum - Rule: $\lim_{x
ightarrow a}(f(x)+g(x))=\lim_{x
ightarrow a}f(x)+\lim_{x
ightarrow a}g(x)$; Difference Rule: $\lim_{x
ightarrow a}(f(x)-g(x))=\lim_{x
ightarrow a}f(x)-\lim_{x
ightarrow a}g(x)$; Product Rule: $\lim_{x
ightarrow a}(f(x)\cdot g(x))=\lim_{x
ightarrow a}f(x)\cdot\lim_{x
ightarrow a}g(x)$; Quotient Rule: $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x
ightarrow a}f(x)}{\lim_{x
ightarrow a}g(x)}$ ($\lim_{x
ightarrow a}g(x)
eq0$); Power Rule: $\lim_{x
ightarrow a}(f(x))^n=(\lim_{x
ightarrow a}f(x))^n$ ($n$ positive integer); Root Rule: $\lim_{x
ightarrow a}\sqrt[n]{f(x)}=\sqrt[n]{\lim_{x
ightarrow a}f(x)}$ ($n$ positive integer, $\lim_{x
ightarrow a}f(x)\geq0$ if $n$ even).

### 2. a.
# Brief Explanations:
The quadratic formula is used to solve quadratic equations of the form $ax^{2}+bx + c = 0$, where $a
eq0$. The solutions for $x$ are given by $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. The expression $b^{2}-4ac$ is called the discriminant, which determines the nature of the roots (real and distinct, real and equal, or complex conjugate).
# Answer:
The quadratic formula for solving $ax^{2}+bx + c = 0$ ($a
eq0$) is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

### 2. b.
# Brief Explanations:
The average rate of change of a function $y = f(x)$ over an interval $[x_1,x_2]$ is the ratio of the change in the function values $\Delta y=f(x_2)-f(x_1)$ to the change in the input values $\Delta x=x_2 - x_1$. It represents the slope of the secant line connecting the points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ on the graph of the function. Mathematically, it is $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$.
# Answer:
The average rate of change of $y = f(x)$ over $[x_1,x_2]$ is $\frac{f(x_2)-f(x_1)}{x_2 - x_1}$, the slope of the secant line between $(x_1,f(x_1))$ and $(x_2,f(x_2))$.

### 2. c.
# Brief Explanations:
The instantaneous rate of change of a function $y = f(x)$ at a point $x = a$ is the limit of the average rate of change of the function over smaller and smaller intervals containing $x = a$. It is equivalent to the slope of the tangent line to the graph of the function at the point $(a,f(a))$ and is calculated as $\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$.
# Answer:
The instantaneous rate of change of $y = f(x)$ at $x = a$ is $\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$, equal to the slope of the tangent line at $(a,f(a))$.

### Part II. 1. a.
# Brief Explanations:
The function $y=\frac{1}{3}$ is a constant function. For any value of $x$, the function value is $\frac{1}{3}$. By the constant - limit rule $\lim_{x
ightarrow c}k = k$ (where $k$ is a constant), when finding $\lim_{x
ightarrow0}\frac{1}{3}$, the limit is $\frac{1}{3}$.
# Answer:
$\frac{1}{3}$

### Part II. 1. b.
# Brief Explanations:
First, substitute $x = 0$ into the function $\sqrt{\frac{x^{2}+16}{2(x + 2)}}$. We get $\sqrt{\frac{0^{2}+16}{2(0 + 2)}}=\sqrt{\frac{16}{4}}=\sqrt{4}=2$. Since the function $\sqrt{\frac{x^{2}+16}{2(x + 2)}}$ is continuous at $x = 0$, $\lim_{x
ightarrow0}\sqrt{\frac{x^{2}+16}{2(x + 2)}} = 2$.
# Answer:
$2$

### Part II. 2. a.
# Brief Explanations:
As $x$ approaches 3 from both the left and the right, the function values approach 4. So $\lim_{x
ightarrow3}f(x)=4$.
# Answer:
$4$

### Part II. 2. b.
# Brief Explanations:
From the graph, when $x = 3$, the function value $f(3)$ is not defined (open - circle), so we write DNE.
# Answer:
DNE

### Part II. 2. c.
# Brief Explanations:
As $x$ approaches 1 from both the left and the right, the function values approach 2. So $\lim_{x
ightarrow1}f(x)=2$.
# Answer:
$2$

### Part II. 2. d.
# Brief Explanations:
From the graph, when $x = 1$, the function value $f(1)=2$.
# Answer:
$2$

### Part II. 2. e.
# Brief Explanations:
As $x$ approaches 2 from both the left and the right, the function values approach 4. So $\lim_{x
ightarrow2}f(x)=4$.
# Answer:
$4$

### Part II. 2. f.
# Brief Explanations:
As $x$ approaches $- 2$ from the right, the function values approach $-2$. So $\lim_{x
ightarrow - 2^{+}}f(x)=-2$.
# Answer:
$-2$

### Part II. 2. g.
# Brief Explanations:
As $x$ approaches $-2$ from the left, the function values approach $-2$. So $\lim_{x
ightarrow - 2^{-}}f(x)=-2$.
# Answer:
$-2$

### Part II. 2. h.
# Brief Explanations:
From the graph, when $x=-2$, the function value $f(-2)=-2$.
# Answer:
$-2$

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QUESTION IMAGE

Question

Explanation:

1. a.

Answer:

1. b.