📝 Piecewise defined functions examples (16 MCQs)

📖 From Calculus • 1. Basics before calculus • 16 questions available

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What is Piecewise defined functions examples?

Definition:
A piecewise defined function is a function that uses different formulas or expressions for different intervals of its domain, allowing it to model situations with distinct behaviors over various ranges.

Example:
$f(x) = \begin{cases} 2x & \text{if } x < 0 \\ x^2 & \text{if } x \geq 0 \end{cases}$ . For $x = -1$ , $f(-1) = -2$ ; for $x = 2$ , $f(2) = 4$ .

Reason:
This flexibility makes piecewise functions ideal for real-world scenarios like tax brackets, shipping costs, or utility rates, where rules change based on thresholds.

Easy

Medium

Hard

📝 All Piecewise defined functions examples MCQs

Q1. In a piecewise-defined function, what term describes a point where the formula changes?

A.Endpoint

B.Breakpoint ✅

C.Critical point

D.Asymptote

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A breakpoint is the specific x‑value at which the rule defining the function switches from one expression to another. Recognizing breakpoints is essential because they determine where separate pieces of the graph meet or diverge, influencing continuity and differentiability.

Q2. Given the piecewise function \ $f(x)=\\begin{cases}0,& x\\le -1\\\\ \\sqrt{1-x^{2}},& -1, which of the following statements is true about \f(-1)\?$

A.f(-1)= -1

B.f(-1)=0 ✅

C.f(-1)=\\sqrt{2}

D.f(-1) is undefined

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The definition explicitly assigns the value 0 to all x that satisfy \ $x\\le -1\$ . Since \ $-1\$ meets this condition, the function value at \ $-1\$ is 0. This direct substitution avoids any ambiguity that might arise from the other pieces of the definition.

Q3. Compare the continuity at \ $x=1\$ for the function in the previous question with the function \ $g(x)=\\begin{cases}0,& x<1\\\\ 2x-1,& x\\ge 1\\end{cases}\$ . Which statement is correct?

A.Both are continuous at 1

B.Both have a jump discontinuity at 1

C.f is continuous but g is discontinuous at 1 ✅

D.g is continuous but f is discontinuous at 1

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: For \ $f\$ , the right‑hand piece gives \ $f(1)=1\$ and the left‑hand limit approaches 1, so \ $f\$ is continuous. For \ $g\$ , the left limit as \ $x\\to1^{-}\$ is 0 while the right limit (and value) is 1, creating a jump. Hence only \ $f\$ is continuous.

Q4. If a piecewise function includes the segment \ $y=\\sqrt{1-x^{2}}\$ on \ $(-1,1)\$ , what geometric shape does this segment represent?

A.Upper half of a circle radius 1 ✅

B.Lower half of a circle radius 1

C.Parabola opening upward

D.Line segment

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The equation \ $y=\\sqrt{1-x^{2}}\$ describes all points satisfying \ $x^{2}+y^{2}=1\$ with \ $y\\ge0\$ . This is precisely the upper semicircle of a unit circle centered at the origin, spanning from \ $-1\$ to \ $1\$ on the x‑axis.

Q5. For the function \ $f\$ above, what is the limit \ $\\displaystyle \\lim_{x\\to -1^{+}} f(x)\$ ?

A.0 ✅

B.1

C.-1

D.Does not exist

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Approaching \ $-1\$ from the right means using the middle piece \ $\\sqrt{1-x^{2}}\$ . As \ $x\$ gets arbitrarily close to \ $-1\$ from the right, the expression under the square root tends to \ $1-(-1)^{2}=0\$ , so the limit equals \ $\\sqrt{0}=0\$ .

Q6. Consider two piecewise functions: \ $f\$ as defined above and \ $h(x)=\\begin{cases}0,& x\\le -1\\\\ 1-x^{2},& -1. Which of the following best describes the difference between their middle pieces?$

A.Both are linear functions

B.f’s middle piece is a semicircle while h’s is a downward parabola ✅

C.Both are constant functions

D.h’s middle piece is undefined

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The middle piece of \ $f\$ is \ $\\sqrt{1-x^{2}}\$ , which traces the upper half of a unit circle—a curved, non‑linear shape. In contrast, \ $h\$ ’s middle piece \ $1-x^{2}\$ is a quadratic expression opening downward, forming a parabola. Thus the two middle pieces have fundamentally different algebraic forms.

Q7. Suppose we define a new piecewise function \ $p(x)\$ that equals the absolute value of the middle piece of \ $f\$ : \ $p(x)=\\begin{cases}0,& x\\le -1\\\\ |\\sqrt{1-x^{2}}|,& -1. How does the graph of \p\ differ from that of \f\ on \(-1,1)\?$

A.It reflects the lower half of the circle above the x‑axis

B.It coincides exactly with \

f\

because the square root is already non‑negative ✅

C.It creates a V‑shape

D.It introduces a gap at x=0

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Since \ $\\sqrt{1-x^{2}}\$ is always non‑negative for \ $-1<x<1\$ , taking its absolute value does not change any y‑values. Consequently, the graph of \ $p\$ on that interval is identical to the original semicircular arc of \ $f\$ .

Q8. If the wind chill index formula \ $W(v)=55.628-22.07v^{0.16}\$ for \ $v>3\$ were mistakenly applied for \ $v\\le 3\$ , which logical error would occur?

A.Using a formula outside its domain ✅

B.Dividing by zero

C.Adding unrelated terms

D.Misidentifying the independent variable

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The piecewise definition of the wind chill function explicitly restricts the empirical formula to speeds greater than 3 mi/h. Applying it to speeds 3 mi/h or less violates the stated domain, leading to results that are not supported by the underlying data. This is a classic domain‑error in logical reasoning.

Q9. Compare the domains of \ $f(x)\$ and the wind chill function \ $W(v)\$ . Which statement is accurate?

A.Both domains are all real numbers

B.Both domains are restricted to non‑negative inputs

C.f’s domain is \

( -\\infty,\\infty)\

while \

W\

’s domain is \

[0,\\infty)\

✅

D.f’s domain is \

[ -1,1]\

only

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The piecewise function \ $f\$ is defined for every real x because each piece covers a portion of the entire real line, giving the domain \ $( -\\infty,\\infty)\$ . The wind chill function, however, is defined only for non‑negative wind speeds, so its domain is \ $[0,\\infty)\$ .

Q10. When graphing a piecewise function, why is it important to plot the function value at each breakpoint separately?

A.To determine the slope of each piece

B.To decide whether the point should be solid or open ✅

C.To compute the area under the curve

D.To find the derivative at that point

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Breakpoints may belong to one piece, both pieces, or none, depending on the definition. Plotting the exact function value at each breakpoint clarifies whether the point is included (solid) or excluded (open) from the graph, which directly affects continuity and visual accuracy.

Q11. Given \ $f\$ as before, if we modify the definition to \ $f_{1}(x)=\\begin{cases}0,& x\\le -1\\\\ \\sqrt{1-x^{2}},& -11\\end{cases}\$ , what logical conclusion follows about the continuity at \ $x=1\$ ?

A.The function becomes discontinuous because the semicircle includes the point (1,0)

B.The function remains continuous because both pieces give the same value at 1 ✅

C.A removable discontinuity appears

D.The function is undefined at 1

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: At \ $x=1\$ , the middle piece yields \ $\\sqrt{1-1^{2}}=0\$ while the right‑hand piece gives \ $x=1\$ . However, the definition of \ $f_{1}\$ assigns the value from the right‑hand piece (since \ $x>1\$ is used) and also includes the point from the middle piece, both yielding the same y‑value of 1. Hence continuity is preserved.

Q12. Analyze the differentiability of \ $f\$ at \ $x=-1\$ . Which of the following is true?

A.f is differentiable at -1 because both side limits of the derivative exist and are equal

B.f is not differentiable at -1 due to a corner point ✅

C.f is differentiable only from the right

D.f is differentiable only from the left

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Approaching \ $-1\$ from the left, \ $f\$ is constant (zero), giving a derivative of 0. Approaching from the right, \ $f\$ follows the semicircle \ $\\sqrt{1-x^{2}}\$ , whose derivative at \ $-1\$ equals \ $\\frac{-x}{\\sqrt{1-x^{2}}}\$ and tends to a finite non‑zero value. The mismatch of one‑sided derivatives creates a corner, so \ $f\$ is not differentiable at \ $-1\$ .

Q13. Synthesize a piecewise function that is continuous everywhere and has a maximum at \ $x=0\$ using the components of \ $f\$ . Which construction achieves this?

A.g(x)=\\begin{cases}\\sqrt{1-x^{2}},& -1\\le x\\le 1\\\\ 0,& \\text{otherwise}\\end{cases} ✅

B.g(x)=\\begin{cases}0,& x\\le -1\\\\ \\sqrt{1-x^{2}},& -1

C.g(x)=\\begin{cases}x^{2},& |x|\\le 1\\\\ 0,& \\text{else}\\end{cases}

D.g(x)=\\begin{cases}1,& x=0\\\\ \\sqrt{1-x^{2}},& -1\\le x\\le 1, x\\neq0\\\\ 0,& \\text{else}\\end{cases}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The function \ $g\$ that retains the upper semicircle on \ $[-1,1]\$ and is zero elsewhere is continuous at the endpoints because the semicircle meets the x‑axis at \ $x=\\pm1\$ . The highest y‑value of the semicircle occurs at \ $x=0\$ where \ $\\sqrt{1-0}=1\$ , satisfying the requirement of a global maximum at the origin.

Q14. If a piecewise function has a breakpoint at \ $x=a\$ where the left-hand limit equals the right-hand limit but the defined value at \ $a\$ differs, what can be inferred about the function at \ $x=a\$ ?

A.It has a jump discontinuity

B.It has a removable discontinuity ✅

C.It is continuous

D.It is undefined

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: When the two one‑sided limits agree, the function approaches a single value from both sides. If the actual assigned value at the breakpoint does not match this common limit, the discontinuity can be removed by redefining the function value to equal the limit. Hence the situation describes a removable discontinuity.

Q15. Compare the effect on the graph when the middle piece of \ $f\$ is replaced by \ $y=|x|\$ on \ $(-1,1)\$ versus the original semicircle. Which statement best describes the change?

A.The graph becomes steeper near the origin

B.The graph retains the same shape but is reflected

C.The graph loses its curvature and becomes V‑shaped ✅

D.The graph becomes a horizontal line

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The original middle piece is a smooth, curved semicircle. Substituting \ $|x|\$ replaces that curvature with two straight lines meeting at the origin, forming a V‑shape. This eliminates the smooth arc and introduces a sharp corner, dramatically altering the visual and analytic properties of the function on that interval.

Q16. Explain why the piecewise definition \ $f(x)=\\begin{cases}0,& x\\le -1\\\\ \\sqrt{1-x^{2}},& -1 is continuous at \x=-1\ but not differentiable there.$

A.Because the left and right limits match but the slopes differ ✅

B.Because the function value is undefined at -1

C.Because the function jumps at -1

D.Because the derivative of the semicircle is infinite at -1

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Approaching \ $-1\$ from the left gives a constant value of 0, while from the right the semicircle also approaches 0, so the limit equals the function value—ensuring continuity. However, the derivative from the left is 0 (flat line) whereas the derivative of \ $\\sqrt{1-x^{2}}\$ approaches a non‑zero value as \ $x\\to-1^{+}\$ . The mismatch of slopes means the derivative does not exist at that point.

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