Increasing and concave down from a derivative sketch

Derivative graph for concavity and monotonicityGiven gradient graph: stationary points (-3,0),(-1.5,-1), origin, x-axis horizontal asymptote and x=1 vertical asymptote. No monotonicity or concavity conclusion is supplied.(−3, 0)(−1.5, −1)x = 1Oy=0y=f′(x)
Question diagram 1
Derivative graph for concavity and monotonicityGiven gradient graph: stationary points (-3,0),(-1.5,-1), origin, x-axis horizontal asymptote and x=1 vertical asymptote. No monotonicity or concavity conclusion is supplied.(−3, 0)(−1.5, −1)x = 1Oy=0y=f′(x)

Given gradient graph with its two turning points, origin and asymptotes.

The diagram below shows the graph of y=f′(x)y=f'(x). The curve passes through the origin and has turning points at (−3,0)(-3,0) and (−1.5,−1)(-1.5,-1). The xx-axis and x=1x=1 are the two asymptotes of the curve.

(i) Find the range of values of xx for which the graph of y=f(x)y=f(x) is strictly increasing and concave downwards. mark

(ii) State the xx-coordinates of all the stationary points of the graph of y=f(x)y=f(x) and determine the nature of each point. marks

[2009/HCI/Prelim/I/3]

Total marks: 3

Solution

(i)

Reading a derivative graph

The graph of y=f′(x)y=f'(x) passes through the origin and has turning points at (−3,0)(-3,0) and (−1.5,−1)(-1.5,-1); the xx-axis and x=1x=1 are asymptotes. Find the range of xx for which y=f(x)y=f(x) is strictly increasing and concave downwards.

Explore this idea

Hint

  1. Strictly increasing means f′>0f'>0: the graph of f′f' is above the xx-axis.

  2. Concave downwards means f′′<0f''<0: the graph of f′f' is falling.

  3. Find where both hold.

Answer · (i)

x>1x>1.

Explore the idea

From f′ back to f

An example of f built from the given gradient graph. The lower graph is the given f′; the upper graph is f.

Position along the curvex = 2
Try an example
Test a critical position

Buttons use full-precision reviewed positions; displayed coordinates are rounded.

This curve is an illustrative reconstruction of the source sketch. Its numerical heights and gradients describe the model; only the reviewed signs and labelled features come from the source.

f(x)
2
Gradient f′(x)
0.6
positive

The gradient is positive and falling: f increases and is concave down. This is the required stretch.

Selected values are within the original drawing windows.

  • An example of f
  • Tangent
  • Given f′, and where f increases
  • Join P to Q; the dashed legs show Δx and Δy
Key idea

Strictly increasing: f′f' above the axis. Concave down: f′f' falling. Both only for x>1x>1.

Ideas to try
Find the stretch where the tangent is rising yet the curve bends downward.
(ii)

Nature of stationary points

State the xx-coordinates of all the stationary points of y=f(x)y=f(x) and determine the nature of each.

Explore this idea

Hint

  1. Stationary points: f′(x)=0f'(x)=0.

  2. Check whether f′f' changes sign.

Answer · (ii)

x=−3x=-3: stationary inflection; x=0x=0: local minimum.

Explore the idea

Does f′ change sign?

Zoom on the two zeros of the gradient graph.

Teal: x = −3 (touches). No comparison overlay.

Start: x = −3 (touches)

f′f' rises to 0 and falls again: it is negative on both sides. No sign change: stationary point of inflexion.

  • Gradient graph f′
  • A zero of f′
Ideas to try
Why does touching the axis without crossing mean no turning point?