12 multiple-choice questions, progressively harder.
Use the function
G(x)={x+1if x<0x2if 0≤x≤27−xif x>2G(x) = \begin{cases} x + 1 & \text{if } x < 0 \\ x^2 & \text{if } 0 \le x \le 2 \\ 7 - x & \text{if } x > 2 \end{cases}G(x)=⎩⎨⎧x+1x27−xif x<0if 0≤x≤2if x>2
to evaluate G(0)G(0)G(0).
Solution
Correct answer: A
The boundary 000 satisfies 0≤x≤20 \le x \le 20≤x≤2, so the middle piece x2x^2x2 applies.
G(0)=02=0G(0) = 0^2 = 0G(0)=02=0
The first piece x+1x + 1x+1 needs x<0x < 0x<0.
H(x)={5if x≤−1x2−1if −1<x<22x−1if x≥2H(x) = \begin{cases} 5 & \text{if } x \le -1 \\ x^2 - 1 & \text{if } -1 < x < 2 \\ 2x - 1 & \text{if } x \ge 2 \end{cases}H(x)=⎩⎨⎧5x2−12x−1if x≤−1if −1<x<2if x≥2
to evaluate H(−3)H(-3)H(−3).
Correct answer: C
Since −3≤−1-3 \le -1−3≤−1, the first piece is the constant 555.
H(−3)=5H(-3) = 5H(−3)=5
to evaluate G(2)G(2)G(2).
Correct answer: D
The boundary 222 satisfies 0≤x≤20 \le x \le 20≤x≤2, so the middle piece applies.
G(2)=22=4G(2) = 2^2 = 4G(2)=22=4
The third piece 7−x7 - x7−x would give 555, but it needs x>2x > 2x>2.
to evaluate H(2)H(2)H(2).
Correct answer: B
The boundary 222 satisfies x≥2x \ge 2x≥2, so the third piece 2x−12x - 12x−1 applies.
H(2)=2(2)−1=3H(2) = 2(2) - 1 = 3H(2)=2(2)−1=3
to evaluate G(3)G(3)G(3).
Since 3>23 > 23>2, the third piece 7−x7 - x7−x applies.
G(3)=7−3=4G(3) = 7 - 3 = 4G(3)=7−3=4
f(x)={x+5if x<0−x+5if x≥0f(x) = \begin{cases} x + 5 & \text{if } x < 0 \\ -x + 5 & \text{if } x \ge 0 \end{cases}f(x)={x+5−x+5if x<0if x≥0
to evaluate f(−4)f(-4)f(−4).
Since −4<0-4 < 0−4<0, the first piece x+5x + 5x+5 applies.
f(−4)=−4+5=1f(-4) = -4 + 5 = 1f(−4)=−4+5=1
This matches 5−∣−4∣=15 - \lvert -4 \rvert = 15−∣−4∣=1.
Does the graph of
jump or connect at x=0x = 0x=0?
Compare both pieces at 000: the left piece x+1x + 1x+1 heads to 111, the middle piece x2x^2x2 is 000.
1≠01 \ne 01=0
Different heights, so the graph jumps at x=0x = 0x=0.
What is the range of
Since f(x)=5−∣x∣f(x) = 5 - \lvert x \rvertf(x)=5−∣x∣ and ∣x∣≥0\lvert x \rvert \ge 0∣x∣≥0, the output is largest when ∣x∣=0\lvert x \rvert = 0∣x∣=0.
f(x)=5−∣x∣≤5f(x) = 5 - \lvert x \rvert \le 5f(x)=5−∣x∣≤5
So the range is y≤5y \le 5y≤5, with the maximum 555 at x=0x = 0x=0.
For what value of aaa does the function
g(x)={2xif x≤3x+aif x>3g(x) = \begin{cases} 2x & \text{if } x \le 3 \\ x + a & \text{if } x > 3 \end{cases}g(x)={2xx+aif x≤3if x>3
connect (no jump) at x=3x = 3x=3?
Set the two pieces equal at x=3x = 3x=3: the left piece is 2(3)=62(3) = 62(3)=6, the right piece is 3+a3 + a3+a.
3+a=6⇒a=33 + a = 6 \Rightarrow a = 33+a=6⇒a=3
Which statement about piecewise functions is NOT true?
Two pieces connect when they meet at the same height, so a piecewise function need not jump at all.
same height at boundary⇒no jump\text{same height at boundary} \Rightarrow \text{no jump}same height at boundary⇒no jump
So 'every piecewise function must jump at its boundaries' is the false statement.
A cinema charges by age aaa in years, in dollars:
{0if a<38if 3≤a<1312if 13≤a<656if a≥65\begin{cases} 0 & \text{if } a < 3 \\ 8 & \text{if } 3 \le a < 13 \\ 12 & \text{if } 13 \le a < 65 \\ 6 & \text{if } a \ge 65 \end{cases}⎩⎨⎧08126if a<3if 3≤a<13if 13≤a<65if a≥65
What is the price at exactly age 656565?
Age 656565 satisfies a≥65a \ge 65a≥65, so the senior tier applies.
price=6 dollars\text{price} = 6 \text{ dollars}price=6 dollars
f(x)={2x+6if x<−1x−4if x≥−1f(x) = \begin{cases} 2x + 6 & \text{if } x < -1 \\ x - 4 & \text{if } x \ge -1 \end{cases}f(x)={2x+6x−4if x<−1if x≥−1
to evaluate f(−1)f(-1)f(−1).
The boundary −1-1−1 satisfies x≥−1x \ge -1x≥−1, so the second piece x−4x - 4x−4 applies.
f(−1)=−1−4=−5f(-1) = -1 - 4 = -5f(−1)=−1−4=−5
The first piece would give 2(−1)+6=42(-1) + 6 = 42(−1)+6=4, but it needs x<−1x < -1x<−1.
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