Implicit differentiation and inverse trigonometric functions, question 26: UBC MATH 100 practice

Question 26 of 30Challenging3 marks · 5 minTheory UBC Math 100 §4.7CLP-1 §2.12Key results

Let $g(x) = \arctan\!\left(\dfrac{1}{x}\right)$ for $x \neq 0$ . Each graph below is drawn with dashed lines at $y = \dfrac{\pi}{2}$ and $y = -\dfrac{\pi}{2}$ . Which one is the graph of $g$ ?

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Theory UBC Math 100 §4.7 CLP-1 §2.12Key results

Worked solution

Strategy

Use the sketch of y = arctan u: increasing, through the origin, levelling off at π/2 as u → ∞ and at −π/2 as u → −∞, and never reaching either level. Then track u = 1/x on each side of 0.

Working

$\arctan 0 = 0, \qquad \lim_{u \to \infty} \arctan u = \frac{\pi}{2}, \qquad \lim_{u \to -\infty} \arctan u = -\frac{\pi}{2}$
The key features of arctan: tan sends (−π/2, π/2) onto every real number, so arctan takes every real input to a value strictly between −π/2 and π/2.
$\lim_{x \to 0^{+}} g(x) = \frac{\pi}{2}, \qquad \lim_{x \to \infty} g(x) = \arctan 0 = 0$
On the right, 1/x runs from +∞ down to 0, so g starts just below π/2 next to the axis and falls toward 0.
$\lim_{x \to 0^{-}} g(x) = -\frac{\pi}{2}, \qquad \lim_{x \to -\infty} g(x) = 0$
On the left, 1/x runs from 0 down to −∞, so g falls from near 0 toward −π/2. g(0) does not exist, and the graph jumps from −π/2 to π/2 across the axis.
$g(1) = \frac{\pi}{4}, \qquad g(-1) = -\frac{\pi}{4}$
Two points to place the branches: halfway up to the dashed line on the right, halfway down on the left, as g is odd.

Answer

\text{Graph B: decreasing branches from } \frac{\pi}{2} \text{ to } 0 \text{ on the right and from } 0 \text{ to } -\frac{\pi}{2} \text{ on the left}

Check

g(1) = arctan 1 = π/4 ≈ 0.785 and g(−1) = −π/4, halfway between 0 and the dashed lines, and only Graph B passes through both points.

Where this goes wrong

Swapping the sides of the jump. Just right of 0, 1/x is huge and positive, so g is close to π/2; just left of 0, 1/x is huge and negative, so g is close to −π/2.

The idea behind it

The implicit differentiation and inverse trigonometric functions lesson, a figure to move, in a few steps

Try a similar question

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\arctan\!\left(\tan\dfrac{3\pi}{4}\right)

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Implicit differentiation and inverse trigonometric functions practice question 26