Do several rising samples prove a function is increasing everywhere on the interval?
My prediction and reason
Predict whether 1/x is decreasing on its whole domain. Separate pairs on the same side of zero from a pair straddling zero.
Evidence and observations
Test 1/x at ordered pairs (−2, −1), (1, 2) and (−1, 1). Record whether the output order agrees with decreasing behaviour.
For f(x) = 2x² + 1, take arbitrary 0 < x₁ < x₂. Factor f(x₂) − f(x₁) and write the sign of every factor.
For −x² + 4x + 1, complete the square and record where the maximum is attained. Explain why a few sampled values cannot rule out an unseen larger value.
Explain or give a counterexample
Transfer
Prove that h(x) = 5 − 3x is strictly decreasing on the reals by comparing arbitrary x₁ < x₂. State the sign of the difference before giving the conclusion.
Exit task
A student says “1/x decreases on its entire domain because each branch slopes down.” Give a counterexample and state the two correct decreasing intervals.
5 min predict the reciprocal claim; 12 min find a counterexample and factor differences; 8 min organise a definition proof; 12 min paper reasoning; 3 min exit check.
Do several rising samples prove a function is increasing everywhere on the interval?
Predict whether 1/x is decreasing on its whole domain. Separate pairs on the same side of zero from a pair straddling zero.
Evidence tasks
- Test 1/x at ordered pairs (−2, −1), (1, 2) and (−1, 1). Record whether the output order agrees with decreasing behaviour.
- For f(x) = 2x² + 1, take arbitrary 0 < x₁ < x₂. Factor f(x₂) − f(x₁) and write the sign of every factor.
- For −x² + 4x + 1, complete the square and record where the maximum is attained. Explain why a few sampled values cannot rule out an unseen larger value.
Reference reasoning
Strict increase on an interval means that every pair x₁ < x₂ in that interval satisfies f(x₁) < f(x₂). Finite checks suggest a conjecture; they do not prove this universal statement. For 2x² + 1 and 0 < x₁ < x₂, the difference 2(x₂ − x₁)(x₂ + x₁) is positive, proving increase. For 1/x, the difference −(x₂ − x₁)/(x₁x₂) is negative when both inputs lie on the same side of zero. The pair −1 < 1 gives −1 < 1 in the outputs, refuting decrease across the entire punctured domain.
Prove that the function is increasing on (0, ∞) using the definition.
f(x)=2x2+1,0<x1<x2;f(x1)−f(x2)=2(x1−x2)(x1+x2) - For arbitrary 0 < x₁ < x₂, x₁ − x₂ < 0 and x₁ + x₂ > 0.
- Therefore f(x₁) − f(x₂) < 0, so f(x₁) < f(x₂). This proves strict increase on (0, ∞).
Find and justify the maximum by completing the square.
f(x)=−x2+4x+1=−(x−2)2+5,x∈R - Since (x − 2)² ≥ 0, f(x) ≤ 5 for every real x.
- Equality holds at x = 2. The maximum is 5, attained at 2.
Transfer and exit task
Prove that h(x) = 5 − 3x is strictly decreasing on the reals by comparing arbitrary x₁ < x₂. State the sign of the difference before giving the conclusion.
A student says “1/x decreases on its entire domain because each branch slopes down.” Give a counterexample and state the two correct decreasing intervals.
Take −1 < 1: f(−1) = −1 < f(1) = 1, contradicting decrease. The function is decreasing separately on (−∞, 0) and (0, ∞).
- Uses ordered allowed inputs on opposite sides of zero.
- States each interval separately and excludes zero.
Questions
- What word in the definition cannot be checked by three samples?
- Where do you use the assumption x₁ and x₂ are positive?
Expected responses
- Students may treat several points as a proof. Ask them to write an arbitrary pair and identify the missing sign argument.
- A correct factorisation can still yield a wrong conclusion if the subtraction direction is forgotten.
Confusions
- Do not merge decreasing intervals across a gap in the domain.
- For an extremum, show both the bound and equality at an allowed input.
Board plan
- Take arbitrary x₁ < x₂ in the interval.
- Compare the difference → determine its sign → apply the definition.
- One counterexample refutes a universal claim; samples alone do not prove it.