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Introduction to the derivative

3.4 Higher order derivatives

1 problem · hints, answers and solutions shown beside each one

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q1Stage 3

A function f(x)f(x) satisfies f′(x)<0f'(x)<0 and f′′(x)>0f''(x)>0 over (a,b)(a,b). Which of the following curves below might represent y=f(x)y=f(x)?

Figure from prob_s2.14, line 2

Figure from prob_s2.14, line 2

Figure from prob_s2.14, line 9

Figure from prob_s2.14, line 9

Figure from prob_s2.14, line 16

Figure from prob_s2.14, line 16

Figure from prob_s2.14, line 25

Figure from prob_s2.14, line 25

Figure from prob_s2.14, line 32

Figure from prob_s2.14, line 32

Hint

Only one of the curves could possibly represent y=f(x)y=f(x).

Answer

(ii)

Full solution

Since f′(x)<0f'(x)<0, we need a decreasing function. This only applies to (ii), (iii), and (v). Since f′′(x)>0f''(x)>0, that means f′(x)f'(x) is increasing, so the slope of the function must be increasing. In (v), the slope is constant, so f′′(x)=0f''(x)=0–therefore, it's not (v). In (iii), the slope is decreasing, because near aa the curve is quite flat (f′(x)f'(x) near zero) but near bb the curve is very steeply decreasing (f′(x)f'(x) is a large negative number), so (iii) has a negative second derivative. By contrast, in (ii), the line starts out as steeply decreasing (f′(x)f'(x) is a strongly negative number) and becomes flatter and flatter (f′(x)f'(x) nears 0), so f′(x)f'(x) is increasing–in other words, f′′(x)>0f''(x)>0. So, (ii) is the only curve that has f′(x)<0f'(x)<0 and f′′(x)>0f''(x)>0.

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From the UBC Math 100 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.