12 multiple-choice questions, progressively harder.
Use the function
f(x)={x+4if x<23x−1if x≥2f(x) = \begin{cases} x + 4 & \text{if } x < 2 \\ 3x - 1 & \text{if } x \ge 2 \end{cases}f(x)={x+43x−1if x<2if x≥2
to evaluate f(5)f(5)f(5).
Solution
Correct answer: B
Since 5≥25 \ge 25≥2, the second piece applies.
f(5)=3(5)−1=14f(5) = 3(5) - 1 = 14f(5)=3(5)−1=14
The value 999 would come from the first piece, which needs x<2x < 2x<2.
to evaluate f(−3)f(-3)f(−3).
Correct answer: D
Since −3<2-3 < 2−3<2, the top piece applies.
f(−3)=−3+4=1f(-3) = -3 + 4 = 1f(−3)=−3+4=1
Only the first piece governs inputs below 222.
D(x)={1if x≤02if 0<x≤25if x>2D(x) = \begin{cases} 1 & \text{if } x \le 0 \\ 2 & \text{if } 0 < x \le 2 \\ 5 & \text{if } x > 2 \end{cases}D(x)=⎩⎨⎧125if x≤0if 0<x≤2if x>2
to evaluate D(2)D(2)D(2).
Correct answer: A
The boundary 222 satisfies 0<x≤20 < x \le 20<x≤2, so the middle piece owns it.
D(2)=2D(2) = 2D(2)=2
The third piece needs x>2x > 2x>2, which excludes 222.
to evaluate D(3)D(3)D(3).
Since 3>23 > 23>2, the third piece applies.
D(3)=5D(3) = 5D(3)=5
Only the last piece governs inputs above 222.
to evaluate D(−1)D(-1)D(−1).
Since −1≤0-1 \le 0−1≤0, the first piece applies.
D(−1)=1D(-1) = 1D(−1)=1
Every input at or below 000 returns 111.
Which statement best describes a piecewise function?
A piecewise function is a single function whose formula changes with where the input falls in the domain.
f(x)={formula Acondition Aformula Bcondition Bf(x) = \begin{cases} \text{formula A} & \text{condition A} \\ \text{formula B} & \text{condition B} \end{cases}f(x)={formula Aformula Bcondition Acondition B
It is one function, and each piece rules its own part of the domain.
In a piecewise definition, what do the conditions on the right of each piece tell you?
Correct answer: C
Each condition states the set of inputs its formula applies to.
piece=(formula) when (condition on x)\text{piece} = (\text{formula}) \text{ when } (\text{condition on } x)piece=(formula) when (condition on x)
So the condition tells you which formula to use for a given input.
To evaluate a piecewise function at x=ax = ax=a, what should you do first?
You must first locate the input among the conditions.
a→which condition holds?→use that piecea \to \text{which condition holds?} \to \text{use that piece}a→which condition holds?→use that piece
Only after choosing the correct piece do you substitute and compute.
A piece has the condition x<3x < 3x<3. What kind of dot does it get at the boundary x=3x = 3x=3?
A strict condition x<3x < 3x<3 never reaches x=3x = 3x=3; it only approaches it.
x<3⇒x=3 excludedx < 3 \Rightarrow x = 3 \text{ excluded}x<3⇒x=3 excluded
An excluded endpoint is drawn as an open (hollow) dot.
If the two pieces reach the same height at a boundary, the graph there:
When both formulas give the same value at the boundary, the two endpoints land on the same point.
left height=right height⇒connects\text{left height} = \text{right height} \Rightarrow \text{connects}left height=right height⇒connects
The open and closed dots merge, so there is no gap.
The domain of a piecewise function is:
The inputs a piecewise function accepts are exactly those covered by its conditions.
domain=(interval1)∪(interval2)∪⋯\text{domain} = (\text{interval}_1) \cup (\text{interval}_2) \cup \cdotsdomain=(interval1)∪(interval2)∪⋯
So the domain is the union of the pieces' intervals.
The floor function ⌊x⌋\lfloor x \rfloor⌊x⌋ is an example of which kind of function?
The floor holds a constant integer value on each unit interval, then jumps.
⌊x⌋=n for n≤x<n+1\lfloor x \rfloor = n \text{ for } n \le x < n + 1⌊x⌋=n for n≤x<n+1
Constant pieces joined by jumps make a step function.
Reset this practice set?
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