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| 14 Theory slides |
| 8 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Mark always likes to have cans of his favorite soft drink in the refrigerator so that he can enjoy a cold drink whenever he wants.
At the beginning of the week, he had 11 cans in the fridge. Over the week, Mark wrote down each time cans were consumed or added.
A piecewise function f is shown below. For every given value of x, find the correct value of f(x). Round the result to two decimal places, if necessary.
When graphing a piecewise function, each piece must be considered separately. First, a piece is graphed for the values of its domain. Then an end — or ends — of the piece are marked with one of the following.
The first piece is f(x)=4x+6. This is a linear function written in slope-intercept form. The function can be graphed by using the y-intercept and the slope.
Now, since the function is limited to inputs less than or equal to -1, the line will be graphed until it reaches x=-1. Since the inequality x≤-1 is non-strict, the function is defined for x=-1 and the circle will be closed.
A similar process can be repeated to graph the second piece f(x)=-x+3. First, the line will be graphed.
This piece is defined for values of x greater than -1. This means that the line will be drawn starting at x=-1. Also, since the inequality is strict, the circle will be open.
Finally, the pieces will be graphed together in the same coordinate plane to complete the graph. It is important to pay attention to the limit points, since each value of x must be assigned to only one value of the function.
The second piece of the function is the horizontal line f(t)=15. Also, since the inequality is non-strict, the circle at t=6 will be filled.
Now, the x-intercept can be used to graph the third and final piece. From Part A, it is known that the x-intercept is 7.5.
Finally, the graph of the piecewise function is completed.
As a summer activity, Tearrik participates in charity events for his community. He is volunteering for a food drive event this weekend.
He went to a shopping center multiple times over the week to collect boxes of food for the food drive. On each day, he collected the following number of boxes.
f(x)=⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧0,5,8,10,13,17,if 0≤x<1if 1≤x<2if 2≤x<3if 3≤x<4if 4≤x<5if 5≤x<6
Day | Collected Boxes | Boxes in Total | x | f(x) |
---|---|---|---|---|
Sunday | - | 0 | 0≤x<1 | 0 |
Monday | 5 | 0+5=5 | 1≤x<2 | 5 |
Tuesday | 3 | 5+3=8 | 2≤x<3 | 8 |
Wednesday | 2 | 8+2=10 | 3≤x<4 | 10 |
Thursday | 3 | 10+3=13 | 4≤x<5 | 13 |
Friday | 4 | 13+4=17 | 5≤x<6 | 17 |
Since step functions are piecewise functions, to graph them, each piece must be considered separately. First, each piece is graphed in its domain as a part of a horizontal line. The end or ends are then marked with one of the following.
The first piece is defined over the interval 0≤x<2. This piece will be graphed by drawing a horizontal line at y=1 from x=0 to x=2.
It can be noted that x=0 is included in the domain of the first piece, but x=2 is not. Therefore, the left end of the segment will be marked with a closed circle and the right end with an open circle.
Then, the same process is repeated for each piece of the function.
Zain is working as a server in a restaurant for a week before their summer vacation ends.
Most of their payment comes from the tips they receive. Zain made note of how much they received in tips, starting from Monday and through their last day working on Saturday.
Day | Tips Received | Total Tips | x | f(x) |
---|---|---|---|---|
Sunday | - | 0 | 0≤x<1 | 0 |
Monday | 49.50 | 0+49.50=49.50 | 1≤x<2 | 49.50 |
Tuesday | 38.17 | 49.50+38.17=87.67 | 2≤x<3 | 87.67 |
Wednesday | 41.45 | 87.67+41.45=129.12 | 3≤x<4 | 129.12 |
Thursday | 58.33 | 129.12+58.33=187.45 | 4≤x<5 | 187.45 |
Friday | 60.55 | 187.45+60.55=248.00 | 5≤x<6 | 248.00 |
Saturday | 57.00 | 248.00+57.00=305.00 | 6≤x<7 | 305.00 |
The next piece is the horizontal segment at 49.50 with a closed endpoint at x=1 and an open endpoint at x=2.
The remaining pieces can be added to the graph by following the same process.
Considering more values can help understand how to draw the graph of the function.
x | f(x)=⌊x⌋ |
---|---|
-1 | -1 |
-0.75 | -1 |
-0.5 | -1 |
-0.25 | -1 |
0 | 0 |
0.5 | 0 |
1 | 1 |
1.25 | 1 |
1.75 | 1 |
1.99 | 1 |
2 | 2 |
From the table above, it can be seen that the function only changes its value when a new integer is reached. It can be noted that the greatest integer function is a step function. Its graph is presented as follows.
Dominika is going to a movie at a local theater on her last day of vacation.
The cost to park in the theater lot is $10 for less than an hour. An additional $2.50 is charged for each hour of parking.
f(x)=2.5⌊x⌋+10
Next, since the greatest integer function is multiplied by 2.5, each value of y is multiplied by 2.5, vertically stretching the spaces between each horizontal segment.
Then, since 10 is added to the product, each segment is translated vertically 10 units up.
Finally, the scope of the coordinate plane will be adjusted so that more steps of the graph can be seen.
During his summer vacation, Ignacio went to private math lessons. After learning about the greatest integer function, Ignacio was asked by his math tutor to graph the numbers y greater than or equal to ⌊x⌋.
Graph the greatest integer function. Then, determine the solution set by shading the appropriate region.
The first step to graph an inequality is to graph the border function. The border function of the given inequality is given by the greatest integer function. This function is a step function whose output is the greatest integer less than or equal to the input x. Note that the inequality is non-strict, so the horizontal lines are drawn as solid lines.
Finally, the following is the complete graph of the inequality that the teacher asked for.
At the beginning of this lesson, it was asked that a function for the number of soda cans in Mark's refrigerator be written. Mark starts the week with 11 cans in the fridge and then does the following.
Dominika plans to go to the waterfall this afternoon after walking through the woods. The path that she will follow is a bit bumpy and, after the wood end, it consists of one hill with a flat area before and after it. Because of this path, her speed will change along the way, but Dominika knows she will reach the waterfall in half an hour.
At 4:00PM, 13 minutes have passed since Dominika left home. Therefore, to know how far away Dominika is, we need to evaluate d(t) at t=13. Since 13 is between 7 and 15, we need to use the second piece of the function. d(t) = 81t, &if 0 ≤ t < 7 60t + 147, &if 7 ≤ t ≤ 15 87t - 258, &if 15 < t < 23 81t - 120, &if 23 ≤ t ≤ 30 We are ready to find out how far from home Dominika was at 4:00PM.
To determine how far away the waterfall is, we need to know how long it takes Dominika to get there. However, rememeber that Dominika said she would reach the waterfall in half an hour. Therefore, we need to evaluate d(t) at t=30 minutes. To do this, we will use the last part of the function. d(t) = 81t, &if 0 ≤ t < 7 60t + 147, &if 7 ≤ t ≤ 15 87t - 258, &if 15 < t < 23 81t - 120, &if 23 ≤ t ≤ 30 Let's find out how far from Dominika's house the waterfall is.
Despite we are not specified how the path is, we do know it consists of one hill preceded and followed by flat regions. Therefore, Dominika should walk at three different speeds.
Consequently, these three speeds should appear in the function rule. To identify them, note that d(t) is formed by four linear functions, and the slope of a linear function can be interpreted as the speed of a moving object. Therefore, let's identify Dominika's speed at each time interval in a table.
Function | Interval | Slope or Speed (m/min) |
---|---|---|
81t | 0 ≤ t < 7 | 81 |
60t+147 | 7 ≤ t ≤ 15 | 60 |
87t-258 | 15 < t < 23 | 87 |
81t-120 | 23 ≤ t ≤ 30 | 81 |
Taking into account that walking uphill slows the speed down, and using the information in the table, let's pair each speed to the corresponding part of the path. cccl Speed & & Path & [0.15cm] 60 m/min & & Uphill &↗ 81 m/min & & Flat & → 87 m/min & & Downhill & ↘ Consequently, Dominika walked uphill during the second time interval. That is, she walked uphill for 15-7 = 8 minutes.
Let f(x) be a piecewise function that consists of two linear functions. The graph of the function is shown.
We start by noticing that the graph of f(x) consists of two pieces. The left-hand piece is defined for x≤ 1 and the right-hand piece is defined for x >1.
Since 16>1, we need to use the right-hand piece to evaluate f(16). However, the value for x=16 is not shown in the graph and we do not know the function rule. Therefore, let's first find the equation of the right-hand piece. To do so, we can use the point-slope form equation. y-y_1 = m(x-x_1) Let's highlight the slope and one point of the right-hand piece.
Now we are ready to find the equation of the right-hand piece.
We have found the equation of the right-hand piece. Now we can proceed with evaluating f(x) at x=16.
Now we have to evaluate f(x) at x=-5. Since -5 < 1, we have to use the left-hand piece. Just like in Part A, we do not know its equation, but we can find it in a similar fashion. However, this time we will use the slope-intercept form equation.
The slope of the line is -2 and the y-intercept is -1. We can use this information to write the equation of the line. y = mx+b ⇓ y = -2x-1 Using this equation and the one we found in Part A, we can write the complete function rule defining f(x). f(x) = -2x-1, &ifx≤ 1 12x + 32, &ifx>1 Finally, let's evaluate f(x) at x=-5.
We are given two piecewise functions, each involving the greatest integer function. Recall that the floor function ⌊ x ⌋ assigns the largest integer that is less than or equal to the value of x. Let's start by identifying the four numbers at which we have to evaluate f(x) and g(x). f( ↑-0.75) + f( ↑2.98)* g( ↑7) - g( ↑-3.05) Since -0.75 is less than 1, we have to use the first piece of f(x) to evaluate it.
Next, let's find f( 2.98). Since 2.98> 1, we will use the second part of f(x).
Now, let's evaluate g(x) at x= 7. Since 7 is greater than -2, we have to use the second part of g(x).
Last but not least, we will find g(-3.05). Here, we will use the first part of g(x) because -3.05 is less than -2.
Finally, let's substitute the values we found into the given expression and simplify.