Justification: Writing the Sentence That Scores
- State what a calculus justification must contain to earn credit
- Write justifications for extrema, concavity, and theorem applications
- Recognize common justifications that read as complete and earn nothing
Why this is its own lesson
On the free-response section, a large share of available points are justification points, and they are lost by students who computed everything correctly. The pattern is consistent: the student finds x = 2, states that there is a maximum there, and moves on — and the reader, who is scoring against a rubric that requires a stated reason, awards the computation and not the conclusion. A justification has to name which derivative, what it does (sign, or sign change), and where, and then draw the conclusion. Three elements, one sentence. Learning the template is worth more points per minute of study than any other topic in the course.
The templates
Local maximum: "f′ changes from positive to negative at x = 2, so f has a local maximum there." Local minimum: negative to positive. Absolute extremum on a closed interval: "Comparing f at the critical points and endpoints, f(0) = 1 is the largest value, so the absolute maximum on [−1, 3] is 1." The comparison itself is the justification, so show the candidate values. Concave up: "f″ > 0 on (2, 5), so f is concave up there." Inflection point: "f″ changes sign from negative to positive at x = 2, so f has an inflection point there" — and the words changes sign are non-negotiable. Increasing: "f′ > 0 on the interval, so f is increasing." Notice every template names a derivative and its behavior; none of them says "because the graph looks like that."
Theorem justifications: state the hypotheses
When a question asks you to justify a conclusion using a theorem, the rubric requires you to verify its hypotheses explicitly. For the IVT: "f is continuous on [1, 4], f(1) = −2 and f(4) = 5, and 0 lies between −2 and 5, so by the Intermediate Value Theorem there is a c in (1, 4) with f(c) = 0." For the MVT: "f is continuous on [0, 6] and differentiable on (0, 6), so by the Mean Value Theorem there is a c in (0, 6) with f′(c) equal to the average rate of change [f(6) − f(0)]/6 = 4." A response that names the theorem and states the conclusion without checking continuity or differentiability typically earns partial credit at best, and the check is one clause.
Given f′(x) = (x − 1)(x − 4)², justify whether f has a local extremum at x = 1 and at x = 4.
- 1.The zeros of f′ are x = 1 and x = 4, so both are critical points.
- 2.Test the sign of f′ around x = 1. At x = 0: (−1)(16) = −16 < 0. At x = 2: (1)(4) = 4 > 0.
- 3.f′ changes from negative to positive at x = 1, so f has a local minimum there.
- 4.Test the sign of f′ around x = 4. At x = 3: (2)(1) = 2 > 0. At x = 5: (4)(1) = 4 > 0.
- 5.f′ is positive on both sides of x = 4, so f′ does not change sign there and f has no extremum at x = 4 — the squared factor is what prevents the sign change.
Justifications that earn nothing, seen constantly: "because f′ = 0 there" (true of maxima, minima and neither); "because the graph turns around" (describes rather than justifies); "because it is the highest point" (restates the conclusion). Each of these accompanies correct arithmetic and still scores zero on the justification point.
Which is an adequate justification that f has a local maximum at x = 3?
To justify by the Mean Value Theorem that some c in (0, 6) has f′(c) = 4, you must state that
Answer the 2 checkpoints as you read.
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