Calculus I, Fall 2021

Brief solutions to Quiz 4

Nov 16, 2021

1. (24 + 16 pts) Write down domains and ranges of all six inverse trigonometric functions. Derive the formula for (cos^{−1})′ (pay attention to the ′+′ or ′−′ sign in your formula and explain).

Ans:

Domains and ranges: See page 3 of Lecture 11 note, or page 62 and page 201 of the textbook (24 pts).

Let θ = cos^{−1}(x),

\frac{d}{dx} cos^{−1}(x) = \frac{1}{\frac{d}{dθ} cos(θ)} = \frac{1}{− sin θ} (4pts)

Since sin^{2} θ + cos^{2} θ = 1 and θ ∈ [0, π], it follows that sin θ ≥ 0 therefore sin θ = +\sqrt{1 − x^{2}} (8 pts).

Answer: \frac{d}{dx} cos^{−1}(x) = \frac{−1}{\sqrt{1 − x^{2}}} (4 pts)

2. (15 + 15 pts) Use the linear approximation formula for (1 + x)^{k} near x = 0 to get an

approximate value of \frac{1}{\sqrt{0.99}} and give an estimate of the error for this approximation

(i.e. find out | true value − approximation | ≤ ⋯). Need not derive the formula you use.

Ans:

\frac{1}{\sqrt{0.99}} = (1 − 0.01)^{\frac{−1}{2}} ≈ (1 − 0.01\frac{−1}{2}) = 1.005

\frac{1}{\sqrt{0.99}} = f(x) = (1 + x)^{\frac{−1}{2}} , x = −0.01

|\frac{1}{\sqrt{0.99}} − 1.005| ≤ \frac{1}{2} max_{−0.01≤c≤0} |f′′(c)| · 0.01^{2}

f′′(c) = \frac{−1}{2} · \frac{−3}{2}(1 + c)^{\frac{−5}{2}}

therefore

|\frac{1}{\sqrt{0.99}} − 1.005| ≤ \frac{1}{2}|f′′(−0.01)| · 0.01^{2} = \frac{3}{8} · 0.99^{\frac{−5}{2}} · 0.01^{2}

3. (15 + 15 pts) Let

f(x) = \{ \begin{matrix}x^{2} sin(\frac{1}{x}), & x ≠ 0, \\ 0, & x = 0. \end{matrix}

Is f differentiable at x = 0? Is f twice differentiable at x = 0? Start with definitions of f′(0) and f′′(0) and explain.

Ans:

f′(0) = lim_{x→0} \frac{x^{2} sin(\frac{1}{x}) − 0}{x − 0} = 0

f′(x) = 2x sin(\frac{1}{x}) − cos(\frac{1}{x}), x ≠ 0

lim_{x→0} f′(x) does not exist

So, f′ is not continuous at x = 0, therefore not differentiable at x = 0 and

f′′(0) = lim_{x→0} \frac{f′(x) − f′(0)}{x − 0} does not exist