For \( f(x) = x – \arctan x \):

\[ f(1) = 1 – \arctan 1 = 1 – \frac{\pi}{4} \]

\[ f(-\sqrt{3}) = -\sqrt{3} – \arctan(-\sqrt{3}) = -\sqrt{3} – \left(-\frac{\pi}{3}\right) = -\sqrt{3} + \frac{\pi}{3} \]

Answer: \( f(1) = 1 – \frac{\pi}{4} \), \( f(-\sqrt{3}) = -\sqrt{3} + \frac{\pi}{3} \).

Compute \( f(-x) \):

\[ f(-x) = -x – \arctan(-x) \]

Since \( \arctan(-x) = -\arctan x \):

\[ f(-x) = -x – (-\arctan x) = -x + \arctan x \]

\[ = -(x – \arctan x) = -f(x) \]

Thus, \( f(-x) = -f(x) \).

Answer: \( f(-x) = -f(x) \).

The range of \( \arctan x \) is \( -\frac{\pi}{2} < \arctan x < \frac{\pi}{2} \) for all \( x \in \mathbb{R} \).

Thus:

\[ -\frac{\pi}{2} < -\arctan x < \frac{\pi}{2} \]

Add \( x \):

\[ x – \frac{\pi}{2} < x – \arctan x < x + \frac{\pi}{2} \]

Since \( f(x) = x – \arctan x \):

\[ x – \frac{\pi}{2} < f(x) < x + \frac{\pi}{2} \]

Add \( \frac{\pi}{2} \):

\[ x – \frac{\pi}{2} + \frac{\pi}{2} < f(x) + \frac{\pi}{2} < x + \frac{\pi}{2} + \frac{\pi}{2} \]

\[ x < f(x) + \frac{\pi}{2} < x + \pi \]

Thus, \( x – \frac{\pi}{2} < f(x) + \frac{\pi}{2} \).

Answer: \( x – \frac{\pi}{2} < f(x) + \frac{\pi}{2} \).

Compute derivatives of \( f(x) = x – \arctan x \):

\[ f'(x) = \frac{d}{dx}(x) – \frac{d}{dx}(\arctan x) = 1 – \frac{1}{1 + x^2} = \frac{x^2}{1 + x^2} \]

\[ f”(x) = \frac{d}{dx} \left( \frac{x^2}{1 + x^2} \right) = \frac{2x(1 + x^2) – x^2 \cdot 2x}{(1 + x^2)^2} = \frac{2x + 2x^3 – 2x^3}{(1 + x^2)^2} = \frac{2x}{(1 + x^2)^2} \]

Evaluate at \( x = 0 \):

\[ f'(0) = \frac{0^2}{1 + 0^2} = 0 \]

\[ f”(0) = \frac{2 \cdot 0}{(1 + 0^2)^2} = 0 \]

Since \( f”(0) = 0 \), check higher derivatives:

\[ f”'(x) = \frac{d}{dx} \left( \frac{2x}{(1 + x^2)^2} \right) = \frac{2(1 + x^2)^2 – 2x \cdot 2(1 + x^2) \cdot 2x}{(1 + x^2)^4} = \frac{2(1 + x^2)^2 – 8x^2(1 + x^2)}{(1 + x^2)^4} \]

\[ = \frac{2(1 + x^2)(1 + x^2 – 4x^2)}{(1 + x^2)^4} = \frac{2(1 – 3x^2)}{(1 + x^2)^3} \]

\[ f”'(0) = \frac{2(1 – 0)}{(1 + 0)^3} = 2 \neq 0 \]

Analyze behavior:

\[ f'(x) = \frac{x^2}{1 + x^2} \geq 0 \text{ for all } x \in \mathbb{R} \]

So, \( f \) is increasing, and \( (0, 0) \) is not a maximum or minimum (since \( f'(x) = 0 \) only at \( x = 0 \)).

\[ f”(x) = \frac{2x}{(1 + x^2)^2} \]

\[ f”(x) > 0 \text{ for } x > 0, \quad f”(x) < 0 \text{ for } x < 0 \]

Since \( f”(x) \) changes sign at \( x = 0 \) and \( f”'(0) \neq 0 \), \( (0, 0) \) is a point of inflection with zero gradient.

Answer: \( f'(x) = \frac{x^2}{1 + x^2} \), \( f”(x) = \frac{2x}{(1 + x^2)^2} \), the graph has a point of inflection at \( (0, 0) \).

Sketch the graph of \( f(x) = x – \arctan x \):

– At \( x = 0 \): \( f(0) = 0 – \arctan 0 = 0 \).

– Asymptotes: As \( x \to \pm \infty \), \( \arctan x \to \pm \frac{\pi}{2} \), so \( f(x) \approx x – \frac{\pi}{2} \) (as \( x \to \infty \)) and \( f(x) \approx x + \frac{\pi}{2} \) (as \( x \to -\infty \)).

– Concavity: From part (d), \( f”(x) > 0 \) for \( x > 0 \) (concave up), \( f”(x) < 0 \) for \( x < 0 \) (concave down).

– Point of inflection: \( (0, 0) \).

Answer: The graph has asymptotes at \( y = x – \frac{\pi}{2} \) and \( y = x + \frac{\pi}{2} \), is concave down for \( x < 0 \), concave up for \( x > 0 \), with a point of inflection at \( (0, 0) \).

Justify that \( f^{-1} \) is defined for all \( x \in \mathbb{R} \):

– From part (d), \( f'(x) = \frac{x^2}{1 + x^2} \geq 0 \), equal to 0 only at \( x = 0 \), so \( f \) is strictly increasing (one-to-one).

– Range: As \( x \to \infty \), \( f(x) \to \infty \); as \( x \to -\infty \), \( f(x) \to -\infty \). Since \( f \) is continuous, its range is \( \mathbb{R} \).

Thus, \( f^{-1} \) exists for all \( x \in \mathbb{R} \).

Sketch: The graph of \( y = f^{-1}(x) \) is the reflection of \( y = f(x) \) over the line \( y = x \). Asymptotes of \( f^{-1} \): \( y = x – \frac{\pi}{2} \) becomes \( x = y – \frac{\pi}{2} \), or \( y = x + \frac{\pi}{2} \), and similarly for the other asymptote.

Answer: \( f^{-1} \) is defined for all \( x \in \mathbb{R} \) because \( f \) is strictly increasing and has range \( \mathbb{R} \). Its graph is the reflection of \( f \) over \( y = x \).