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

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

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

n = 5 or n = −4

1. (a) b = 4, c = 9
2. (b) b = −32, c = 38

1. (a) Minimum at \(\left(\frac{25}{16},\ - \frac{25}{8}\right)\)
2. (b) Minimum at \(\left(-2,\ - \frac{1}{4}\right)\); maximum at \(\left(2,\ \frac{1}{4}\right)\)
3. (c) Maximum at \(\left(\frac{\pi}{6},\ \frac{\pi}{6} + \sqrt{3}\right)\); minimum at \(\left(\frac{5 \pi}{6},\ - \sqrt{3} + \frac{5 \pi}{6}\right)\)

1. (a) y″ = 6 > 0
2. (b) y″ = −4y
3. (c) y″ = −4/(a + 2x)² < 0

1. (a) x < 3
2. (b) (3, −5)

1. (a) x = 0.750
2. (b) x = 0.558

1. (a) \(f'(x)=\dfrac{1-3\ln x}{x^4}\)
2. (b) \(\left(e^{1/3},\ \dfrac{1}{3e}\right)\)
3. (c) \(f''(x)=\dfrac{12\ln x-7}{x^5}\)
4. (d) Local maximum

1. (a) x = 7
2. (b) P = 68 hundred dollars

t = 2 hours, rate = 5 litres per hour

1. (a) dP/dt = 216t/(9+t²)²; t = 1.73 years
2. (b) P = 3.00 thousand; greatest rate = 2.60 thousand per year

Minimum at \(\left(\dfrac{2}{\sqrt p},\ 4\sqrt p\right)\); maximum at \(\left(-\dfrac{2}{\sqrt p},\ -4\sqrt p\right)\)

1. (a) \(x=\frac{a}{3}\)
2. (b) \(V=\frac{a^{3}}{27}\)

1. (a) \(f(x)\ge\dfrac{a}{2}\left(1-\ln\dfrac{a}{2}\right)\)
2. (b) \(U_{\min}=\dfrac{5}{3}\left(\dfrac{3}{2}\right)^{2/5}\) J

1. (a) b > 3
2. (b) b/a = 9

\(y=x^{3} + 2 x^{2} - 2 x + 4\)

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