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Worked solution

Reconstruct a decreasing function with an isolated zero gradient

The diagram shows y=f′(x)y=f^{\prime}(x) meeting the x-axis at x=−3x=-3 and x=1x=1. Given f(−3)=4f(-3)=4 and f(1)=1f(1)=1:

(i) State the coordinates and natures of the stationary points of y=f(x)y=f(x). marks

(ii) Find the range of xx for which the curve is strictly decreasing. marks

(iii) Sketch y=f(x)y=f(x). marks

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Given derivative is negative before −3, touches zero at −3 as a local maximum, falls to an unlabelled negative minimum, then rises and crosses the axis at 1. It is positive after 1.
Given derivative is negative before −3, touches zero at −3 as a local maximum, falls to an unlabelled negative minimum, then rises and crosses the axis at 1. It is positive after 1.
Given derivative is negative before −3, touches zero at −3 as a local maximum, falls to an unlabelled negative minimum, then rises and crosses the axis at 1. It is positive after 1.

[2010/TPJC/I/6]

CJC · Review problems · Question 5

Answer

Stationary inflection (−3,4)(-3,4); minimum (1,1)(1,1). Printed negative-gradient ranges exclude −3-3, but the function strictly decreases through it on (−∞,1)(-\infty,1).

(i) Classify the two labelled stationary points

(−3,4)(-3,4) is a stationary point of inflection: f′f^{\prime} touches zero from below at −3-3 and changes from increasing to decreasing. (1,1)(1,1) is a minimum, since f′f^{\prime} changes from negative to positive at 11.

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(ii) Compare negative gradients with strict monotonicity

The printed answer gives x<−3x<-3 or −3<x<1-3<x<1, the ranges where f′<0f^{\prime}<0. There is an isolated zero at −3-3, not a constant interval. Thus, as a statement about the function’s strict monotonicity, ff is strictly decreasing throughout (−∞,1)(-\infty,1), including through its stationary inflection at −3-3. A derivative may vanish at an isolated point of a strictly decreasing function.

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(iii) Retain both source coordinate anchors

The source function decreases from the upper left, flattens through the stationary inflection (−3,4)(-3,4), steepens again, and reaches the minimum (1,1)(1,1) before increasing. It stays above the x-axis. The exact y-intercept and the position of the other inflection implied by the unlabelled derivative minimum are not numerically supplied; the figure is schematic, not a uniquely determined function.

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Compare your sketch

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Function sketch decreases through stationary inflection (−3,4), then reaches minimum (1,1) and increases. Both points are above the x-axis, and the whole function shown stays above it. No numerical y-intercept is specified.
Function sketch decreases through stationary inflection (−3,4), then reaches minimum (1,1) and increases. Both points are above the x-axis, and the whole function shown stays above it. No numerical y-intercept is specified.
Function sketch decreases through stationary inflection (−3,4), then reaches minimum (1,1) and increases. Both points are above the x-axis, and the whole function shown stays above it. No numerical y-intercept is specified.

Common mistakes

One isolated zero derivative does not break strict decrease.

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