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Written Assignment 1

Question 1

1. What linear function, y=f(x) has f(0) = 8 and f(7) = 14 ?

\[ y = mx + b \text{ --------(1)} \\ 14 = m(7) + b \text{ --------(2)} \\ 8 = m(0) + b \text{ --------(3)} \\ \text{ } \\ \text{From (2) and (3), we find m} \\ \text{ } \\ m = \frac{14-8}{7-0} = \frac{6}{7} \\ \text{} \\ \text{substitute m into (2)} \\ 14 = \frac{6}{7}(7) + b \\ 14 = 6 + b \\ b = 8 \\ \text{ } \\ \text{substitute m and b into (1) which is the final solution} \\ \text{} \\ y = \frac{6}{7}x + 8 \\ \text{ } \\ \]

Question 2

2. If \(f(t)=2t-t^2+3\) , what is \(\frac{f(t+h)-f(t)}{h}\) ?

Let’s find f(t+h)

\[ \text{ } \\ f(t+h) = 2(t+h) - (t+h)^2 + 3 \\ f(t+h)= 2t + 2h - (t^2 + 2th + h^2) + 3 \\ f(t+h)= 2t + 2h - t^2 - 2th - h^2 + 3 \\ \text{ } \\ \]

And then substitute f(t+h) and f(t) into the equation

\[ \begin{align} \frac{f(t+h)-f(t)}{h } &= \frac{2t + 2h - t^2 - 2th - h^2 + 3 - (2t-t^2+3)}{h} \\ &= \frac{2t + 2h - t^2 - 2th - h^2 + 3 - 2t + t^2 - 3}{h} \\ &= \frac{2h - 2th - h^2}{h} \\ &= \frac{h(2 - 2t - h)}{h} \\ &= 2 - 2t - h \\ \end{align} \]

Question 3

3. Find all solutions to the equation \(-4cosx=-sin^2x+1\). Write your answer in radians in terms of \(\pi\)

we know that \(sin^2x = 1 - cos^2x\), so we can use it in the equation

\[ \begin{align} -4cosx &= -sin^2x + 1 \\ -4cosx &= -(1 - cos^2x) + 1 \\ -4cosx &= -1 + cos^2x + 1 \\ -4cosx &= cos^2x \\ cos^2x + 4cosx &= 0 \\ cosx(cosx + 4) &= 0 \\ \end{align} \]

The equation is true when \(cosx = 0\) or \(cosx = -4\)

  • \(x=-4\) is refused as it is not in the domain of \(cosx\) which is \([-1,1]\).
  • \(x=0\) is accepted.

So for \(cos(x) = 0\) we have \(x = \frac{\pi}{2} + \pi n\) where \(n\) is an integer.

Question 4

4. Sketch the graph of \(y=3^{x+2} +5\). Find the domain, range, and horizontal asymptote. Include the horizontal asymptote in your graph.

  • The domain is \((-\infty, \infty)\) as there is no restriction on the value of \(x\).
  • The range is \((5, \infty)\). as the term \(3^{x+2}\) is always positive and \(5\) is added to it.
  • The line \(y=5\) is the horizontal asymptote (colored blue in the graph below) as \(x\) approaches \(-\infty\) the value of \(y\) approaches \(5\).

Question 5

5. Sketch the graph of \(y= log\_ 3(x+2) +5\). Find the domain, range, and vertical asymptote. Include the vertical asymptote in your graph.

  • The domain is \((-2, \infty)\) as \(x+2\) must be positive (as it is the argument of the logarithm function).
  • The range is \((-\infty, \infty)\) as the logarithm function can take any value.
  • The line \(x=-2\) is the vertical asymptote (colored green in the graph below) as \(x\) approaches \(-2\) the value of \(y\) approaches \(-\infty\).

Question 6

6. Find the domain fo the function \(g(x)= \frac{2x}{x^2-81}\).

The function is defined for all values of \(x ∈ \mathbb{R}\) except when the denominator is zero. So we need to find the values of \(x\) for which \(x^2-81=0\) which is solved for \(x = \pm 9\).

So the domain is \((-\infty, -9) \cup (-9, 9) \cup (9, \infty)\) or \(\mathbb{R} - \{-9, 9\}\).

Question 7

7. From the graph below, find what is f(-15) and for what numbers x is f(x)=0.

  • \(f(-15) = 0\) as the graph crosses the x-axis at \(x=-15\).
  • \(f(x)=0\) when \(x=-15\) and \(x=17\) as the graph crosses the x-axis at \(x=-15\) and \(x=17\).

Question 8

8. Determine whether the graph is that of a function. If it is, use the graph to find its domain and range, the intercepts, if any, and any symmetry with respect to the x-axis, the y-axis, or the origin.

  • The graph is function as for each value of \(x\) there is only one value of \(y\).
  • The domain is \((-\infty, \infty)\) as there is no restriction on the value of \(x\).
  • The range is \([-1, 1]\) as the value of \(y\) is always between -1 and 1.
  • The graph crosses the x-axis 5 times each period. So the x-intercepts in the period shown in the graph are \(x=-\frac{3}{2}, -\frac{1}{2}, 0, \frac{1}{2}, \frac{3}{2}\).
  • Symmetry with respect to the y-axis as the graph is symmetric about the y-axis and origin.
  • No symmetry with respect to the x-axis.

Question 9

9. Determine whether the function is even, odd, or neither.

a. \(f(x)=4x^3\) b. \(f(x)= \frac{-x^3}{4x^2+3}\) c. \(f(x)=3x^3-5\)

A function is even if \(f(-x)=f(x)\) and odd if \(f(-x)=-f(x)\)..

a. \(f(x)=4x^3\) is odd.

\[ \begin{align} f(-x) &= 4(-x)^3 \\ &= - 4x^3 \\ &= -f(x) \\ \end{align} \]

b. \(f(x)= \frac{-x^3}{4x^2+3}\) is odd.

\[ \begin{align} f(-x) &= \frac{-(-x)^3}{4(-x)^2+3} \\ &= \frac{-(-x)^3}{4x^2+3} \\ &= - \frac{-x^3}{4x^2+3} \\ &= -f(x) \\ \end{align} \]

c. \(f(x)=3x^3-5\) is neither.

\[ \begin{align} f(-x) &= 3(-x)^3 - 5 \\ &= -3x^3 - 5 \\ &≠ -f(x) \\ &≠ f(x) \\ \end{align} \]

Question 10

10. A cellular phone plan had the following schedule of charges: Basic service, including 100 minutes of calls is $20.00/month; 2nd 100 minutes of calls is $0.075/minute; additional minutes of calls is $0.10/minute.

a. What is the charge for 200 minutes of calls in one month?

  • The first 100 minutes are included in the basic service ($20.00/month).
  • The next 100 minutes are charged at \(0.075/minute (\)0.075 × 100 = 7.50$).
  • Thus, the charge for 200 minutes is \(20.00 + 7.50 = 27.50\).

b. What is the charge for 250 minutes of calls in one month?

  • The first 100 minutes are included in the basic service ($20.00/month).
  • The next 100 minutes are charged at \(0.075/minute (\)0.075 × 100 = 7.50$).
  • The next 50 minutes are charged at \(0.10/minute (\)0.10 × 50 = 5.00$).
  • Thus, the charge for 250 minutes is \(20.00 + 7.50 + 5.00 = 32.50\).

c. Construct a function that relates the monthly charge C for x minutes of calls?

\[ C(x) = \begin{cases} 20, & \text{if } x \leq 100 \\ 20 + 0.075(x - 100), & \text{if } 100 < x \leq 200 \\ 20 + 0.075 \times 100 + 0.10(x - 200), & \text{if } x > 200 \end{cases} \]