Mixed practice: stationary points, concavity, parameters and models Problem set B Answers

A curve that always rises. It bends up (concave up) for x < −1 and bends down (concave down) for x > −1, with a point of inflection at x = −1.

1. (a) p = −18
2. (b) Point of inflection at x = 3: the curve changes from concave down to concave up.

n = 6 or n = −5

1. (a) b = −8, c = 17
2. (b) b = −4, c = 6

1. (a) Minimum at \(\left(\frac{16}{9},\ - \frac{16}{3}\right)\)
2. (b) Minimum at \(\left(-3,\ - \frac{1}{6}\right)\); maximum at \(\left(3,\ \frac{1}{6}\right)\)
3. (c) Maximum at \(\left(\frac{2 \pi}{3},\ \sqrt{3} + \frac{2 \pi}{3}\right)\); minimum at \(\left(\frac{4 \pi}{3},\ - \sqrt{3} + \frac{4 \pi}{3}\right)\)

1. (a) y″ = 8 > 0
2. (b) y″ = −25y
3. (c) y″ = −9/(a + 3x)² < 0

1. (a) x < −2
2. (b) (−2, 25)

1. (a) x = 0.829
2. (b) x = 0.638

1. (a) \(f'(x)=\dfrac{1-4\ln x}{x^5}\)
2. (b) \(\left(e^{1/4},\ \dfrac{1}{4e}\right)\)
3. (c) \(f''(x)=\dfrac{20\ln x-9}{x^6}\)
4. (d) Local maximum

1. (a) x = 8
2. (b) P = 54 thousand dollars

t = 3 minutes, rate = 3 litres per minute

1. (a) dP/dt = 640t/(16+t²)²; t = 2.31 months
2. (b) P = 5.00 thousand; greatest rate = 3.25 thousand per month

Minimum at \(\left(\dfrac{3}{\sqrt p},\ 6\sqrt p\right)\); maximum at \(\left(-\dfrac{3}{\sqrt p},\ -6\sqrt p\right)\)

1. (a) \(x=\frac{2 a}{9}\)
2. (b) \(V=\frac{4 a^{3}}{243}\)

1. (a) \(f(x)\ge\dfrac{a}{3}\left(1-\ln\dfrac{a}{3}\right)\)
2. (b) \(U_{\min}=4\left(\dfrac{1}{3}\right)^{3/4}\) J

1. (a) b > 16/3
2. (b) b/a = 27/2

\(y=2 x^{3} - 3 x^{2} + 3 x - 3\)

1. (a) Local minimum at x = 3. There is no local maximum.
2. (b) x = −1 and x = 5/3
3. (c) x = −1 only