Calculus I, Fall 2014 (http://www.math.nthu.edu.tw/~wangwc/)

Solutions in Quiz03

1. If f(x) is continuous on [a, b] and differentiable on (a, b), then there is one point c in (a, b) such that

f′(c) = \frac{f(b) − f(a)}{b − a}

2. Absolute maximum is f(−3) = 6\sqrt[3]{9} Absolute minimum is f(3) = f(0) = 0.

3. It’s true. Let f(x) = cos x, then f(x) is continuous and differentiable on R. For all values of x, by Mean Value Theorem,

\frac{cos x − cos 0}{x − 0} = sin c

for some c ∈ (0, x).

Hence

| cos x − 1| ≤ |x|

for all values of x.

4. Let f(x) = e^{x} − 1 − x, then f(0) = 0 and f′(x) = e^{x} − 1 ≥ 0 on x ≥ 0.

Hence f(x) ≥ f(0) = 0 on x ≥ 0. That is e^{x} − 1 − x ≥ 0 on x ≥ 0.

Let g(x) = e^{x} − 1 − x − \frac{x^{2}}{2} , then g(0) = 0 and g′(x) = e^{x} − 1 − x ≥ 0 on x ≥ 0.

Hence g(x) ≥ g(0) = 0 on x ≥ 0. That is e^{x} ≥ 1 + x + \frac{x^{2}}{2} on x ≥ 0.

5. f′(x) = 4x^{3} − 12x^{2} = 4x^{2}(x − 3) and f′′(x) = 12x^{2} − 24x = 12x(x − 2). We have

\{ \begin{matrix}Critical points: x=0, 3 \\ Points of inflection: x=0, 2 \end{matrix}

and f(x)

Decreasing and concave up on (−∞, 0] Decreasing and concave down on [0, 2] Decreasing and concave up on [2, 3] Increasing and concave up on [3, ∞)