Porisma

Original question & working

Recover the signs hidden by an absolute value · source reference

2021/DHS/Promo/7 · Question pp. 3 · Working pp. 7

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The working and teaching notes are included in the interactive lesson.

A minimum of |f| need not be a minimum of f. Closed endpoints describe full branches, not strict derivative inequalities.

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Full original question, working & notes

Printed question

The graphs of y=f′(x)y=f'(x) and y=∣f(x)∣y=|f(x)| are shown below. The gradient graph has turning point (−1,−2)(-1,-2), intercepts OO and (3,0)(3,0), and asymptotes y=0y=0, x=2x=2. The absolute-value graph shows (0,1)(0,1), (1,0)(1,0), (3,3)(3,3) and asymptotes y=0y=0, x=2x=2.

(i) State the nature of all stationary points of the graph of y=f(x)y=f(x). [2]

(ii) State the range of values of xx where (a) ff is decreasing, [2] (b) ff is concave upwards. [1]

(iii) Sketch the graph of y=f(x)y=f(x) indicating clearly the asymptotes, coordinates of the stationary points and the intersections with the axes. [3]

[2021/DHS/Promo/7]

Printed working

(i) Maximum point (3,−3)(3,-3), minimum point (0,−1)(0,-1).

(ii)(a) Printed interval: x≤0x\le0 or x≥3x\ge3. (ii)(b) Printed interval: −1≤x<2-1\le x<2.

(iii) Reconstructed ff is negative for x<1x<1, crosses at (1,0)(1,0), positive for 1<x<21<x<2, and negative for x>2x>2, with extrema (0,−1)(0,-1) and (3,−3)(3,-3). Asymptotes are y=0y=0 (to the left) and x=2x=2.

Endpoint qualification

For strict derivative sign statements use f′<0f'<0 on x<0x<0 or x>3x>3, and f′′>0f''>0 on −1<x<2-1<x<2. The printed closed endpoints express decreasing/convex interval conventions; derivative is zero at the displayed boundaries. The discontinuity at 22 splits the increasing ranges.

Teacher notes

To sketch y=f(x)y=f(x):

  1. Curve y=∣f(x)∣y=|f(x)| is obtained from y=f(x)y=f(x) by reflecting the part where y<0y<0 above the xx-axis.

  2. To obtain y=f(x)y=f(x) reflect some parts of y=∣f(x)∣y=|f(x)| below the xx-axis.

3(a). From y=f′(x)y=f'(x), xx-intercept (0,0)(0,0) becomes a stationary point at x=0x=0 for y=f(x)y=f(x).

3(b). f′(x)>0f'(x)>0 for 0<x<30<x<3, x≠2x\ne2: the gradient of y=f(x)y=f(x) is positive, so ff is strictly increasing on the separate intervals 0<x<20<x<2 and 2<x<32<x<3. Reflect ∣f(x)∣|f(x)| where 0<x<10<x<1 and 2<x<32<x<3 below the axis.

3(c). f′(x)<0f'(x)<0 for x<0x<0 or x>3x>3: the gradient of y=f(x)y=f(x) is negative, so ff is strictly decreasing there. Reflect ∣f(x)∣|f(x)| where x<0x<0 and x>3x>3 below the axis.

Answer key

(i) Max (3,−3)(3,-3), min (0,−1)(0,-1). (ii)(a) x≤0x\le0 or x≥3x\ge3; (ii)(b) −1≤x<2-1\le x<2. (iii) Same native reconstructed sketch.

Native diagrams

Original gradient graph, including (−1,−2), O, (3,0) and asymptotes. Original |f| graph including (0,1), (1,0), (3,3). Reconstructed f with all original answer labels.