12 multiple-choice questions, progressively harder.
Use the function
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(−1)H(-1)H(−1).
Solution
Correct answer: D
The boundary −1-1−1 satisfies x≤−1x \le -1x≤−1, so the first piece applies.
H(−1)=5H(-1) = 5H(−1)=5
The middle piece needs x>−1x > -1x>−1, so it excludes −1-1−1.
Does the graph of
jump or connect at x=−1x = -1x=−1?
Compare both pieces at −1-1−1: the left piece is 555, the middle piece x2−1x^2 - 1x2−1 heads to (−1)2−1=0(-1)^2 - 1 = 0(−1)2−1=0.
5≠05 \ne 05=0
Different heights, so the graph jumps at x=−1x = -1x=−1.
What is the range of
Correct answer: C
Piece by piece: the constant 555; then x2−1x^2 - 1x2−1 on (−1,2)(-1, 2)(−1,2) gives [−1,3)[-1, 3)[−1,3); then 2x−12x - 12x−1 on x≥2x \ge 2x≥2 gives [3,∞)[3, \infty)[3,∞).
{5}∪[−1,3)∪[3,∞)=[−1,∞)\{5\} \cup [-1, 3) \cup [3, \infty) = [-1, \infty){5}∪[−1,3)∪[3,∞)=[−1,∞)
So the range is y≥−1y \ge -1y≥−1.
K(x)={4−x2if x≤04if 0<x<2x+2if x≥2K(x) = \begin{cases} 4 - x^2 & \text{if } x \le 0 \\ 4 & \text{if } 0 < x < 2 \\ x + 2 & \text{if } x \ge 2 \end{cases}K(x)=⎩⎨⎧4−x24x+2if x≤0if 0<x<2if x≥2
to evaluate K(−2)K(-2)K(−2).
Correct answer: A
Since −2≤0-2 \le 0−2≤0, the first piece 4−x24 - x^24−x2 applies.
K(−2)=4−(−2)2=4−4=0K(-2) = 4 - (-2)^2 = 4 - 4 = 0K(−2)=4−(−2)2=4−4=0
jump or connect at x=0x = 0x=0?
Correct answer: B
Compare both pieces at 000: 4−02=44 - 0^2 = 44−02=4 and the constant 444.
4=44 = 44=4
Same height, so the graph connects at x=0x = 0x=0.
jump or connect at x=2x = 2x=2?
Compare both pieces at 222: the constant 444 and 2+2=42 + 2 = 42+2=4.
Same height, so the graph connects at x=2x = 2x=2.
to evaluate K(3)K(3)K(3).
Since 3≥23 \ge 23≥2, the third piece x+2x + 2x+2 applies.
K(3)=3+2=5K(3) = 3 + 2 = 5K(3)=3+2=5
Postage in dollars is
{2if 0<w≤14if 1<w≤26if 2<w≤3\begin{cases} 2 & \text{if } 0 < w \le 1 \\ 4 & \text{if } 1 < w \le 2 \\ 6 & \text{if } 2 < w \le 3 \end{cases}⎩⎨⎧246if 0<w≤1if 1<w≤2if 2<w≤3
Find the postage for w=2w = 2w=2 pounds.
The weight 222 satisfies 1<w≤21 < w \le 21<w≤2, so the second tier applies.
postage=4 dollars\text{postage} = 4 \text{ dollars}postage=4 dollars
The boundary 222 belongs to the tier whose condition includes it.
Find the postage for w=1w = 1w=1 pound.
The weight 111 satisfies 0<w≤10 < w \le 10<w≤1, so the first tier applies.
postage=2 dollars\text{postage} = 2 \text{ dollars}postage=2 dollars
Find the postage for w=3w = 3w=3 pounds.
The weight 333 satisfies 2<w≤32 < w \le 32<w≤3, so the third tier applies.
postage=6 dollars\text{postage} = 6 \text{ dollars}postage=6 dollars
A step function has a constant value on each interval, with neighboring values different. Between one interval and the next, its graph:
Each tier is constant and neighboring tiers differ, so the value leaps at each boundary.
⋯→2→4→6→⋯\cdots \to 2 \to 4 \to 6 \to \cdots⋯→2→4→6→⋯
The graph jumps at each boundary, the mark of a step function.
The function
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
can be written as which single formula?
For x≥0x \ge 0x≥0, −x+5=5−x=5−∣x∣-x + 5 = 5 - x = 5 - \lvert x \rvert−x+5=5−x=5−∣x∣; for x<0x < 0x<0, x+5=5−(−x)=5−∣x∣x + 5 = 5 - (-x) = 5 - \lvert x \rvertx+5=5−(−x)=5−∣x∣.
f(x)=5−∣x∣f(x) = 5 - \lvert x \rvertf(x)=5−∣x∣
Both pieces are the same single rule in disguise.
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