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Unit 5 · Topic 5.8 Beta

Sketching f, f' and f''

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So far you have mostly worked from formulas. The exam just as often hands you a picture and asks about a different function: here is ff, describe f′f'; here is f′f', describe ff. You do not need a formula for either. You need to read features, and each feature of one graph becomes a feature of the other.

From ff to f′f': heights become slopes

The height of the graph of the derivative at each xx is the slope of ff at that xx. So read the graph of ff one feature at a time:

Graph of ffGraph of f′f'
risingabove the xx-axis
fallingbelow the xx-axis
a smooth peak or valley (flat tangent line)crosses the xx-axis
concave uprising
concave downfalling
a point of inflectiona peak or valley
a straight line with slope mma horizontal segment at height mm
a cornera jump; no value at that point
Top: f(x) = x cubed over 3 minus x squared minus 3x plus 2, with a peak at x = -1, a valley at x = 3 and an inflection point at x = 1. Bottom: f'(x) = x squared minus 2x minus 3. f' is zero at x = -1 and x = 3, where f turns, and lowest at x = 1, where f changes concavity; f' is positive where f rises and negative where f falls.
Figure 1. Line the graphs up. The peak and valley of ff are the zeros of f′f'; the inflection point of ff is the lowest point of f′f'. LevlPrep original diagram.

Worked example 1. The graph of f(x)=x44−2x2f(x) = \frac{x^4}{4} - 2x^2 is a W shape: valleys at x=−2x = -2 and x=2x = 2, a peak at x=0x = 0, and inflection points near x=±1.15x = \pm 1.15. Sketch f′f'.

Zeros of f′f': at the peak and valleys, x=−2,0,2x = -2, 0, 2.

Signs: ff falls on (−∞,−2)(-\infty, -2) and (0,2)(0, 2), so f′<0f' < 0 there; ff rises on (−2,0)(-2, 0) and (2,∞)(2, \infty), so f′>0f' > 0.

Turns of f′f': at the inflection points of ff. f′f' has a peak near x=−1.15x = -1.15 and a valley near x=1.15x = 1.15.

Check with the formula: f′(x)f'(x)=x3−4x= x^3 - 4x=x(x−2)(x+2)= x(x - 2)(x + 2), a cubic with exactly those zeros and turns.

From f′f' to ff: slopes become shapes

Read the same table backward. Given the graph of f′f':

  • Where f′f' is positive, ff rises; where negative, ff falls.
  • Where f′f' crosses the axis going down, ff has a peak; going up, a valley. Where f′f' touches the axis without crossing, ff has a flat point but keeps going.
  • Where f′f' rises, ff is concave up; where it falls, concave down; where it turns, ff has a point of inflection.
  • Where f′f' is constant, ff is a straight line with that slope.

The graph of f′f' does not tell you how high ff is. Many graphs of ff, shifted up or down, share one derivative. To place ff you need one value, such as f(0)=4f(0) = 4.

Worked example 2. The graph of f′f' is made of segments from (−2,−2)(-2, -2) to (0,2)(0, 2), to (2,2)(2, 2), to (4,−2)(4, -2). Describe ff.

f′f' crosses zero going up at x=−1x = -1 (valley of ff) and going down at x=3x = 3 (peak of ff).

On (−2,0)(-2, 0), f′f' rises: ff is concave up. On (0,2)(0, 2), f′=2f' = 2: ff is a straight segment with slope 2. On (2,4)(2, 4), f′f' falls: concave down.

f″f'' is the slope of the graph of f′f': 22, then 00, then −2-2. At x=0x = 0 and x=2x = 2 the slope of f′f' jumps, so f″f'' does not exist there.

One level further down

The same pattern links f′f' and f″f'': the height of the graph of f″f'' is the slope of f′f'. So from a graph of f′f' you can read f″(3)f''(3) as a slope, and say where f″f'' is positive (wherever f′f' is rising).

A symmetry shortcut

Reflecting a graph across the yy-axis reverses every slope. So if ff is even (symmetric about the yy-axis, like x2+cos⁡xx^2 + \cos x), then f′f' is odd (symmetric about the origin), and the reverse holds too.

Common sketching mistakes

  • Copying the shape of ff. The graph of f′f' does not look like ff shifted; peaks of ff are zeros of f′f'.
  • Putting the peaks of f′f' at the peaks of ff. The peaks of f′f' are the steepest points of ff, its inflection points.
  • Smoothing over a corner. A corner in ff is a jump in f′f'.

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