Calculus I, Fall 2013

Brief solutions to Midterm Exam 1

1. (8 pts) Find lim_{y→+∞}y sin \frac{2}{\sqrt{y}}.

Ans:

= lim_{y→+∞}\sqrt{y} 2 \frac{\sqrt{y}}{2} sin \frac{2}{\sqrt{y}} = ∞ · 2 · 1 = ∞.

2. (12 pts) Give precise definition of lim_{x→0^{+}} \frac{1}{x} = ∞ and show that it is true (using the ε − δ argument). Ans: For any M ∈ R, there exist δ > 0 such that 0 < x < δ implies f(x) > M. Proof: If M > 0, take δ = 1/M. If M ≤ 0, take any δ > 0 (for example δ = 1) will do. Then verify (details skipped).

3. (8 pts) Find dy/dx where y = x^{x}, x > 0. Need not simplify your expression. Ans: = x^{x}(1 + ln x).

4. (12 pts) Find y′ and y′′ at (1, −1) where y(x) is implicitly given by tan(x+y)+sin(x^{2} + y) = 0. Ans: \frac{d}{dx} once:

( sec^{2}(x + y))(1 + y′) + ( cos(x^{2} + y))(2x + y′) = 0,

evaluate at x = 1, y = −1, one gets y′ = −3/2. \frac{d}{dx} twice: (2 sec^{2}(x+y) tan(x+y))(1+y′)^{2}+( sec^{2}(x+y))y′′−( sin(x^{2}+y))(2x+y′)^{2}+( cos(x^{2}+y))(2+y′′) = 0 evaluate at x = 1, y = −1, y′ = −3/2, one gets y′′ = −1.

5. (12 pts) Find the smallest n such that \frac{d^{n}}{dx^{n}} (x^{10} sin x)|_{x=0} is nonzero and find this value.

Ans: n = 11. \frac{d^{11}}{dx^{11}} (x^{10} sin x)|_{x=0} = 11!.

6. (12 pts) True or False? If true, prove it. If false, give a counter example. If |f(x) − (3x + 2)| ≤ |x|^{1.5} for all x ∈ R, then f is differentiable at x = 0. Ans: True. Step 1: let x = 0, we have f(0) = 2. Step 2:

lim_{x→0} \frac{f(x) − f(0)}{x − 0} = lim_{x→0} \frac{f(x) − (3x + 2)}{x − 0} + 3.

Since |\frac{f(x)−(3x+2)}{x−0}| ≤ |x|^{0.5}, it follows from Sandwich Theorem that

lim_{x→0} \frac{f(x) − (3x + 2)}{x − 0} = 0.

Therefore

lim_{x→0} \frac{f(x) − f(0)}{x − 0} = 3, differentiable.

7. (12 pts) Write down L(x, x_{0}), the linear approximation of f near x_{0}. Find an approximate value of sin(\frac{π}{3} − 0.01) such that the error of the approximation is smaller than 5 × 10^{−5}. (Hint: choose x_{0} carefully) Ans: L(x, x_{0}) = f(x_{0}) + f′(x_{0})(x − x_{0}). Choose x_{0} = \frac{π}{3}. sin(\frac{π}{3} − 0.01) ≈ sin(\frac{π}{3}) + cos \frac{π}{3} (−0.01) = \frac{\sqrt{3}}{2} − 0.005.

|Error| ≤ \frac{1}{2}| sin c| 0.01^{2} ≤ 5 × 10^{−5}

8. (12 pts) True or False? (prove it if true, correct it if false).

Since x ↦ ln x and x ↦ e^{x} are inverse function to each other and \frac{d}{dx} ln x = \frac{1}{x}. Therefore \frac{d}{dx}e^{x} = \frac{1}{\frac{d}{dx} ln x} = \frac{1}{\frac{1}{x}} = x. Ans: False. Correction: \frac{d}{dx}e^{x} = \frac{1}{( \frac{d}{dy} ln y)_{y=e^{x}}} = \frac{1}{( \frac{1}{y})_{y=e^{x}}} = e^{x}.

9. (12 pts)

Find absolute maximum and absolute minimum of f(x) = x^{1/3}(x − 1/2) on [−1, 1]. Ans: f′(x) = \frac{1}{6}x^{\frac{−2}{3}} (8x − 1). x = 0, f is not differentiable. x = 1/8, f′(1/8) = 0. End points: x = −1, 1. Compare f(x) on x = −1, 1, 0, 1/8, we find that absolute maximum = f(−1) = 3/2, absolute minimum = f(1/8) = −3/16.