Read increase and curvature from derivative signs

Try

A curve is above the horizontal axis but moving downward. Is its derivative positive because its value is positive, or negative because it is decreasing? Sketch a positive-height segment with negative slope. Then draw one that is decreasing but bending upward: slope and concavity describe different features.

Predict

For f(x)=x³−3x, predict whether every zero of f′ must be a global extremum. The polynomial grows without bound in opposite directions at its two ends, so a global maximum or minimum on all real inputs is unlikely. Local turning behavior is a separate question.

Experiment

Factor f′(x)=3x²−3=3(x−1)(x+1), then test signs between its zeros. Compute f″(x)=6x independently. Use the two charts to build a sketch instead of tracing a computer plot.

Interval	Sign of f′	Function behavior
x less than −1	positive	increasing
−1<x<1	negative	decreasing
x greater than 1	positive	increasing

Interval	Sign of f″	Concavity
x negative	negative	down
x positive	positive	up

The local maximum is (−1,2), the local minimum is (1,−2), and the inflection point is (0,0). Mark these points along with the sign information. A local maximum here has a larger value than its neighbors, not than every point on the full graph.

Understand

If f′ is positive throughout an interval, the mean value theorem establishes that f is increasing there; a negative derivative establishes decrease. The first derivative test classifies a critical point by signs around it: positive to negative indicates a local maximum, negative to positive a local minimum. A derivative zero without sign change does not establish an extremum, as x³ at zero shows.

Concavity describes how slopes change. Positive f″ indicates slopes increasing and concavity upward; negative f″ indicates slopes decreasing and concavity downward, on intervals where these conditions hold. An inflection point is a point of the graph where concavity changes. A zero of f″ is only a candidate. For x⁴, the second derivative 12x² is zero at zero but positive on both sides, so there is no concavity change there.

At a stationary point, the second derivative test gives a local minimum if f″ is positive and a local maximum if it is negative, under the usual differentiability conditions near the point. If f″ is zero, the test is inconclusive. The examples x⁴, −x⁴ and x³ at zero respectively have a minimum, a maximum and neither, while all have first and second derivatives zero there. An inconclusive test is not proof that no extremum exists.

Domain checks come before chart conclusions. For 1/x, the second derivative 2/x³ changes sign across zero, but zero is not a point of the graph. It is not an inflection point. Its first derivative is negative on both domain intervals, yet you cannot apply an uninterrupted-interval theorem across the excluded input. Asymptotes and holes belong on the sketch alongside critical points.

Graph reconstruction should also identify intercepts, endpoint behavior and known symmetry when useful. A sign chart controls direction and bending but does not determine every numerical height. Two vertically shifted functions share their derivative graphs, so derivative information alone needs an anchor value to determine vertical placement.

Apply

In the practice bank, justify every classification by a sign change or another valid argument. For the x⁴ error repair, use nonnegativity or the first derivative test rather than repeating an inconclusive second derivative test. Distinguish a missing point from a zero derivative at an existing point.

Continue

The next lesson chooses global extrema on a specified feasible domain. Local shape analysis supplies candidates, but boundaries, nondifferentiable points and existence conditions determine whether the design comparison is complete.