Let's solve the problem **step-by-step** for \( f(x) = 1 + \frac{1}{x} - \frac{1}{x^2} \): --- ### (a) **Vertical and Horizontal Asymptotes** **Vertical Asymptotes:** Set denominator equal to zero (undefined points): \[ x = 0 \] So, **vertical asymptote** at \( x = 0 \). --- **Horizontal Asymptote:** Take the limit as \( x \to \pm\infty \): \[ \lim_{x \to \infty} f(x) = 1 + 0 - 0 = 1 \] \[ \lim_{x \to -\infty} f(x) = 1 + 0 - 0 = 1 \] So, **horizontal asymptote** at \( y = 1 \). --- ### (b) **Intervals of Increase or Decrease** Find the derivative: \[ f(x) = 1 + x^{-1} - x^{-2} \] \[ f'(x) = -x^{-2} + 2x^{-3} = -\frac{1}{x^2} + \frac{2}{x^3} = \frac{-x + 2}{x^3} \] Set \( f'(x) = 0 \): \[ \frac{-x + 2}{x^3} = 0 \implies -x + 2 = 0 \implies x = 2 \] Check sign of \( f'(x) \) in intervals \((-\infty, 0)\), \((0, 2)\), \((2, \infty)\): - For \( x < 0 \): numerator (-), denominator (-) → \(+\) (increasing) - For \( 0 < x < 2 \): numerator (+), denominator (+) → \(+\) (increasing) - For \( x > 2 \): numerator (-), denominator (+) → \(-\) (decreasing) So, **increasing on \( (-\infty, 0) \cup (0, 2) \)** **Decreasing on \( (2, \infty) \)** --- ### (c) **Local Maximum and Minimum Values** From above, \( x=2 \) is a critical point. - \( f'(x) \) changes from positive to negative at \( x = 2 \), so **local maximum** at \( x = 2 \). Calculate value: \[ f(2) = 1 + \frac{1}{2} - \frac{1}{4} = 1 + 0.5 - 0.25 = 1.25 \] No local minimum in the domain besides behavior at infinity. --- ### (d) **Intervals of Concavity and Inflection Points** Second derivative: \[ f'(x) = \frac{-x + 2}{x^3} \] \[ f''(x) = \frac{-1 \cdot x^3 - (-x + 2) \cdot 3x^2}{x^6} = \frac{-x^3 - 3x^2(-x + 2)}{x^6} = \frac{-x^3 + 3x^3 - 6x^2}{x^6} = \frac{2x^3 - 6x^2}{x^6} = \frac{2x^2(x - 3)}{x^6} = \frac{2(x - 3)}{x^4} \] Set \( f''(x) = 0 \): \[ 2(x - 3) = 0 \implies x = 3 \] Check sign of \( f''(x) \): - For \( x < 0 \): numerator (-), denominator (+) → \(-\) (concave down) - For \( 0 < x < 3 \): numerator (-), denominator (+) → \(-\) (concave down) - For \( x > 3 \): numerator (+), denominator (+) → \(+\) (concave up) **Inflection point** at \( x = 3 \): \[ f(3) = 1 + \frac{1}{3} - \frac{1}{9} = 1 + 0.333 - 0.111 = 1.222 \] --- ### (e) **Sketch the Graph** **Summary:** - Vertical asymptote: \( x = 0 \) - Horizontal asymptote: \( y = 1 \) - Increasing: \( (-\infty, 0) \cup (0, 2) \) - Decreasing: \( (2, \infty) \) - Local max: \( (2, 1.25) \) - Inflection point: \( (3, 1.222) \) - Concave down: \( (-\infty, 3) \) - Concave up: \( (3, \infty) \) --- #### **Rough Sketch** - For \( x \to 0^- \), \( f(x) \to -\infty \) - For \( x \to 0^+ \), \( f(x) \to +\infty \) - For large \(|x|\), \( f(x) \to 1 \) - Maximum at \( x = 2 \), inflection at \( x = 3 \) --- #### **Final Answers** **(a)** Vertical asymptote: \( x = 0 \); Horizontal asymptote: \( y = 1 \) **(b)** Increasing: \( (-\infty, 0) \cup (0, 2) \); Decreasing: \( (2, \infty) \) **(c)** Local maximum at \( x = 2 \) with value \( 1.25 \) **(d)** Concave down on \( (-\infty, 3) \), up on \( (3, \infty) \); inflection at \( x = 3 \) **(e)** See above for graphing details! (You can plot these features for a rough sketch.)

Let's carefully analyze the function \( f(x) = 1 + \frac{1}{x} - \frac{1}{x^2} \) step-by-step, including detailed calculations, to answer all parts. --- ### **Part (a): Find the vertical and horizontal asymptotes** #### Vertical asymptotes: Vertical asymptotes occur where the function is undefined, i.e., where the denominator is zero. Since \( f(x) \) involves \( \frac{1}{x} \) and \( \frac{1}{x^2} \), the denominator is zero at: \[ x = 0 \] **Vertical asymptote:** \[ \boxed{x = 0} \] --- #### Horizontal asymptotes: Evaluate the limits as \( x \to \pm \infty \): \[ \lim_{x \to \infty} f(x) = 1 + \frac{1}{x} - \frac{1}{x^2} \] As \( x \to \infty \), both \( \frac{1}{x} \to 0 \) and \( \frac{1}{x^2} \to 0 \), so: \[ \lim_{x \to \infty} f(x) = 1 + 0 - 0 = 1 \] Similarly, as \( x \to -\infty \): \[ \lim_{x \to -\infty} f(x) = 1 + 0 - 0 = 1 \] **Horizontal asymptote:** \[ \boxed{y = 1} \] --- ### **Part (b): Intervals of increase/decrease** #### Step 1: Compute the derivative \( f'(x) \): Given \[ f(x) = 1 + \frac{1}{x} - \frac{1}{x^2} \] Rewrite as: \[ f(x) = 1 + x^{-1} - x^{-2} \] Differentiate: \[ f'(x) = 0 + (-1) x^{-2} - (-2) x^{-3} = - \frac{1}{x^2} + \frac{2}{x^3} \] Express with common denominator \( x^3 \): \[ f'(x) = - \frac{x}{x^3} + \frac{2}{x^3} = \frac{-x + 2}{x^3} \] #### Step 2: Find critical points (where \( f'(x) = 0 \)): Set numerator zero (denominator cannot be zero): \[ -x + 2 = 0 \implies x = 2 \] Note: \( x = 0 \) is excluded since the function is undefined there. #### Step 3: Sign of \( f'(x) \): - For \( x < 0 \): Numerator: \( -x + 2 \): - At \( x = -1 \): \( -(-1) + 2 = 1 + 2 = 3 > 0 \) Denominator: \( x^3 \): - \( x^3 \) is negative for \( x < 0 \) Therefore: \[ f'(-1) = \frac{\text{positive}}{\text{negative}} < 0 \] Actually, check the sign carefully: - Numerator: \( -x + 2 \). For \( x < 0 \), \( -x > 0 \), so numerator > 0, denominator < 0. Thus, \( f'(x) < 0 \) for \( x < 0 \). - For \( 0 < x < 2 \): Numerator: \( -x + 2 \). Since \( x < 2 \), numerator > 0. Denominator: \( x^3 > 0 \). So, \( f'(x) > 0 \) in this interval. - For \( x > 2 \): Numerator: \( -x + 2 \). Since \( x > 2 \), numerator < 0. Denominator: \( x^3 > 0 \). Therefore, \( f'(x) < 0 \) for \( x > 2 \). --- ### **Summary of increasing/decreasing:** - **Decreasing** on \( (-\infty, 0) \) (since \( f' < 0 \)) - **Increasing** on \( (0, 2) \) (since \( f' > 0 \)) - **Decreasing** on \( (2, \infty) \) (since \( f' < 0 \)) --- ### **Part (c): Local maxima and minima** - At \( x = 2 \), \( f'(x) \) changes from positive to negative, indicating a **local maximum**. Calculate \( f(2) \): \[ f(2) = 1 + \frac{1}{2} - \frac{1}{4} = 1 + 0.5 - 0.25 = 1.25 \] **Answer:** \[ \boxed{ \text{Local maximum at } x=2, \quad f(2) = 1.25 } \] - No local minimum is indicated because \( f' \) does not change from negative to positive. --- ### **Part (d): Concavity and inflection points** #### Step 1: Find second derivative \( f''(x) \): Recall: \[ f'(x) = \frac{-x + 2}{x^3} \] Use quotient rule or differentiate directly: \[ f''(x) = \frac{d}{dx} \left( \frac{-x + 2}{x^3} \right) \] Let numerator \( u = -x + 2 \), denominator \( v = x^3 \): \[ u' = -1, \quad v' = 3x^2 \] Using quotient rule: \[ f''(x) = \frac{u' v - u v'}{v^2} = \frac{(-1)(x^3) - (-x + 2)(3x^2)}{(x^3)^2} \] Simplify numerator: \[ - x^3 - (-x + 2)(3x^2) = -x^3 + (x - 2)(3x^2) \] Distribute: \[ -x^3 + 3x^3 - 6x^2 = ( -x^3 + 3x^3 ) - 6x^2 = 2x^3 - 6x^2 \] Denominator: \[ x^6 \] Thus, \[ f''(x) = \frac{2x^3 - 6x^2}{x^6} = \frac{2x^2 (x - 3)}{x^6} = \frac{2(x - 3)}{x^4} \] --- #### Step 2: Find inflection points (where \( f''(x) = 0 \)): \[ 2(x - 3) = 0 \implies x = 3 \] Note that \( x = 0 \) is excluded (vertical asymptote). #### Step 3: Sign of \( f''(x) \): - For \( x < 0 \): Numerator: \( 2(x - 3) \): - For \( x < 0 \), numerator is negative (since \( x - 3 < 0 \)) - Denominator: \( x^4 > 0 \) Thus, \( f''(x) < 0 \) — **concave down** on \( (-\infty, 0) \). - For \( 0 < x < 3 \): Numerator: \( 2(x - 3) \): - Negative (since \( x < 3 \)) - Denominator: \( x^4 > 0 \) So, \( f''(x) < 0 \) — **concave down** on \( (0, 3) \). - For \( x > 3 \): Numerator: positive (since \( x - 3 > 0 \)) Denominator: positive Thus, \( f''(x) > 0 \) — **concave up** on \( (3, \infty) \). **Inflection point at \( x=3 \):** Calculate \( f(3) \): \[ f(3) = 1 + \frac{1}{3} - \frac{1}{9} = 1 + 0.333... - 0.111... = 1.222... \] --- ### **Part (e): Graph sketching summary** - Vertical asymptote at \( x=0 \): \( f(x) \to -\infty \) as \( x \to 0^- \), and \( f(x) \to +\infty \) as \( x \to 0^+ \). - Horizontal asymptote at \( y=1 \): \( f(x) \to 1 \) as \( x \to \pm \infty \). - Increasing on \( (0, 2) \), decreasing on \( (-\infty, 0) \) and \( (2, \infty) \). - Local maximum at \( (2, 1.25) \). - Inflection point at \( (3, 1.222) \). - Concave down on \( (-\infty, 3) \), concave up on \( (3, \infty) \). --- ## **Final Summary:** | Part | Result | |---------|--------------------------------------------------------------| | (a) | **Vertical asymptote:** \( x=0 \) <br> **Horizontal asymptote:** \( y=1 \) | | (b) | **Increasing:** \( (0, 2) \) <br> **Decreasing:** \( (-\infty, 0) \cup (2, \infty) \) | | (c) | **Local maximum:** at \( x=2 \), \( f=1.25 \) | | (d) | **Concave down:** \( (-\infty, 3) \) <br> **Concave up:** \( (3, \infty) \) <br> **Inflection point:** at \( x=3 \), \( f \approx 1.222 \) | | (e) | Graph features as described above; sketch accordingly. | --- This detailed analysis covers all parts with step-by-step calculations.

**Final Answers:** **(a)** Vertical asymptote at **\( x = 0 \)** Horizontal asymptote at **\( y = 1 \)** **(b)** Increasing on **\( (0, 2) \)** Decreasing on **\( (-\infty, 0) \cup (2, \infty) \)** **(c)** Local maximum at **\( x = 2 \)** with **\( f(2) = 1.25 \)** **(d)** Concave down on **\( (-\infty, 3) \)** Concave up on **\( (3, \infty) \)** Inflection point at **\( x = 3 \)** with **\( f(3) \approx 1.222 \)** **(e)** Graph features: - As \( x \to 0^- \), \( f(x) \to -\infty \) - As \( x \to 0^+ \), \( f(x) \to +\infty \) - \( f(x) \to 1 \) as \( x \to \pm \infty \) - Increases from 0 to 2, with a maximum at \( x=2 \) - Inflection at \( x=3 \), changing from concave down to up.