91Ó°ÊÓ

To find vertical asymptotes, check where the function is undefined. Since \(f(x) = \sqrt{x^2 + 1} - x\) has a square root and subtraction, we note that \(\sqrt{x^2 + 1}\) is defined and positive for all \(x\). Thus, there are no vertical asymptotes. For horizontal asymptotes, evaluate \( \lim_{{x \to \infty}} f(x)\) and \( \lim_{{x \to -\infty}} f(x)\).Calculate: \[ \lim_{{x \to \infty}} \sqrt{x^2+1} - x = \lim_{{x \to \infty}} \frac{x^2 + 1 - x^2}{\sqrt{x^2 + 1} + x} \approx 0, \]since the numerator tends to a constant and the denominator grows very large.Similarly, \[ \lim_{{x \to -\infty}} \sqrt{x^2+1} - x = \lim_{{x \to -\infty}} \frac{x^2 + 1 - x^2}{\sqrt{x^2 + 1} - x} = 1. \]Thus, a horizontal asymptote is at \(y = 0\) for \(x \to \infty\) and \(y = 1\) for \(x \to -\infty\).

To find the intervals of increase or decrease, calculate the derivative \(f'(x)\) and find where it is positive or negative.Derivative calculation:\[ f(x) = \sqrt{x^2 + 1} - x \Rightarrow f'(x) = \frac{x}{\sqrt{x^2 + 1}} - 1. \]Set \(f'(x) = 0\) to find critical points:\[ \frac{x}{\sqrt{x^2 + 1}} - 1 = 0 \Rightarrow x = \sqrt{x^2 + 1}. \]Solving for \(x\), square both sides:\[ x^2 = x^2 + 1 \Rightarrow 1 = 0 \text{ (no real solution)}. \]The function does not have critical points where \(f'(x) = 0\), hence check sign changes of \(f'(x)\) on intervals:- As \(x \to -\infty, f'(x) < 0\)- As \(x \to \infty, f'(x) < 0\)Thus, the function is always decreasing.

Since the derivative does not equal zero anywhere in real numbers, there are no critical points for local maximum or minimum values. The function is decreasing over its entire domain.

Find the second derivative \(f''(x)\) to identify concavity:\[ f'(x) = \frac{x}{\sqrt{x^2 + 1}} - 1 \rightarrow f''(x) = \frac{(x^2+1)^{3/2} - x^2(x^2 + 1)^{-1/2}}{x^2 + 1}, \]simplified further.Due to complexity, observe the concavity from earlier steps:- Function appears concave up as it approaches horizontal asymptotes.- No inflection points detected as neither derivative changes sign.

Using information from parts (a)-(d), sketch the graph:- The graph approaches a horizontal asymptote \(y = 0\) as \(x \to \infty\) and \(y = 1\) as \(x \to -\infty\).- Since the function is always decreasing with no local extrema, it should start higher on the left (near \(y=1\) at negative infinity) and lower on the right (near \(y=0\) at positive infinity).- The sketch should reflect the continuous, smooth decrease in function value as \(x\) progresses.