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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 --------(1)14=m(7)+b --------(2)8=m(0)+b --------(3) From (2) and (3), we find m m=14870=67substitute m into (2)14=67(7)+b14=6+bb=8 substitute m and b into (1) which is the final solutiony=67x+8 

Question 2

2. If f(t)=2tt2+3 , what is f(t+h)f(t)h ?

Let’s find f(t+h)

 f(t+h)=2(t+h)(t+h)2+3f(t+h)=2t+2h(t2+2th+h2)+3f(t+h)=2t+2ht22thh2+3 

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

f(t+h)f(t)h=2t+2ht22thh2+3(2tt2+3)h=2t+2ht22thh2+32t+t23h=2h2thh2h=h(22th)h=22th

Question 3

3. Find all solutions to the equation 4cosx=sin2x+1. Write your answer in radians in terms of π

we know that sin2x=1cos2x, so we can use it in the equation

4cosx=sin2x+14cosx=(1cos2x)+14cosx=1+cos2x+14cosx=cos2xcos2x+4cosx=0cosx(cosx+4)=0

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=π2+πn where n is an integer.


Question 4

4. Sketch the graph of y=3x+2+5. Find the domain, range, and horizontal asymptote. Include the horizontal asymptote in your graph.

  • The domain is (,) as there is no restriction on the value of x.
  • The range is (5,). as the term 3x+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 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,) as x+2 must be positive (as it is the argument of the logarithm function).
  • The range is (,) 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 .


Question 6

6. Find the domain fo the function g(x)=2xx281.

The function is defined for all values of xR except when the denominator is zero. So we need to find the values of x for which x281=0 which is solved for x=±9.

So the domain is (,9)(9,9)(9,) or 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 (,) 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=32,12,0,12,32.
  • 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)=4x3 b. f(x)=x34x2+3 c. f(x)=3x35

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

a. f(x)=4x3 is odd.

f(x)=4(x)3=4x3=f(x)

b. f(x)=x34x2+3 is odd.

f(x)=(x)34(x)2+3=(x)34x2+3=x34x2+3=f(x)

c. f(x)=3x35 is neither.

f(x)=3(x)35=3x35f(x)f(x)

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)={20,if x10020+0.075(x100),if 100<x20020+0.075×100+0.10(x200),if x>200