Recover the signs hidden by an absolute value

Original f′Source-faithful qualitative diagram; distances are schematic unless an exact function is stated.xyO(−1,−2)(3,0)x=2y=0y=f′(x)
Question diagram 1
Original f′Source-faithful qualitative diagram; distances are schematic unless an exact function is stated.xyO(−1,−2)(3,0)x=2y=0y=f′(x)

Given f′ graph with turning point(-1,-2), intercepts and asymptotes.

Original absolute-value graphSource-faithful qualitative diagram; distances are schematic unless an exact function is stated.xyO(0,1)(1,0)(3,3)x=2y=0y=|f(x)|
Question diagram 2
Original absolute-value graphSource-faithful qualitative diagram; distances are schematic unless an exact function is stated.xyO(0,1)(1,0)(3,3)x=2y=0y=|f(x)|

Given absolute-value graph with labelled points and asymptotes.

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). marks

(ii) State the range of values of xx where (a) ff is decreasing, marks (b) ff is concave upwards. mark

(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. marks

[2021/DHS/Promo/7]

Total marks: 8

Solution

(i)

Stationary points from f′f'

The graph of f′f' has a turning point (−1,−2)(-1,-2), intercepts OO and (3,0)(3,0) and asymptotes y=0y=0, x=2x=2. The graph of ∣f∣|f| shows (0,1)(0,1), (1,0)(1,0), (3,3)(3,3) and asymptotes y=0y=0, x=2x=2. State the nature of all stationary points of ff.

Hint

  1. Stationary points are the zeros of f′f': x=0x=0 and x=3x=3.

  2. Check whether f′f' changes from negative to positive (minimum) or positive to negative (maximum).

  3. Use ∣f∣|f| to find the yy-coordinates and signs.

Answer · (i)

Minimum (0,−1)(0,-1); maximum (3,−3)(3,-3).

(ii)(a)

Monotonicity from f′f'

State the range of values of xx where ff is decreasing.

Hint

  1. Decreasing means f′<0f'<0: the graph of f′f' lies below the xx-axis.

Answer · (ii)(a)

Decreasing branches (−∞,0](-\infty,0] and [3,∞)[3,\infty).

The derivative is strictly negative only for x<0 or x>3; stationary endpoints may be included when naming whole branches.

(ii)(b)

Concavity from f′f'

State the range of values of xx where ff is concave upwards.

Hint

  1. Concave up means f′′>0f''>0, so the graph of f′f' is rising.

Answer · (ii)(b)

Concave-up branch [−1,2)[-1,2).

Strict f″>0 only on −1<x<2; f″=0 at −1, and x=2 is excluded.

(iii)

Reconstructing from ∣f∣|f|

Sketch y=f(x)y=f(x), indicating the asymptotes, the stationary points and the intersections with the axes.

Explore this idea

Hint

  1. Start with the graph of ∣f∣|f|.

  2. Decide which parts of ff are negative: where ff is decreasing to a minimum and to a maximum below the axis.

  3. Reflect those parts below the xx-axis.

Reconstructing f from absolute f
Reconstructing f from absolute fScroll across to see the full diagram.

Answer · (iii)

Intercepts (0,−1),(1,0)(0,-1),(1,0); stationary points above; asymptotes x=2x=2, left-end y=0y=0.

Explore the idea

From |f| back to f

An example consistent with the given points. Reflect the parts that were folded upwards.

Teal: y = |f(x)| (given). No comparison overlay.

Start: y = |f(x)| (given)

The given graph: (0,1)(0,1), (1,0)(1,0) and (3,3)(3,3), asymptotes x=2x=2 and y=0y=0.

  • Graph at the selected step
  • Vertical asymptote x = 2
Ideas to try
Which part of the graph does not move when you reflect?