The function f is defined by f(x) = (x^2 + 1)/(x + 1), where x ∈ ℝ, x ≠ -4. For the graph of f: (a) Write down the equation of the vertical asymptote. (b) Find the equation of the... The function f is defined by f(x) = (x^2 + 1)/(x + 1), where x ∈ ℝ, x ≠ -4. For the graph of f: (a) Write down the equation of the vertical asymptote. (b) Find the equation of the horizontal asymptote. (c) Using an algebraic approach, show that the graphs of f and f^(-1) intersect at x = p and x = q, where p < q. (i) Find the value of p and the value of q. (ii) Hence, find the area enclosed by the graph of f and the graph of f^(-1).
Understand the Problem
The question is asking to analyze the function defined in terms of vertical and horizontal asymptotes, find specific values related to the function and its inverse, and determine the area enclosed by the graphs. This involves understanding key concepts of calculus and algebra.
Answer
Vertical asymptote: $x = 4$, Horizontal asymptote: $y = 1$, Inverse: $f^{-1}(y) = \frac{4y + 4}{y - 1}$, No area is enclosed.
Answer for screen readers
- Vertical asymptote: $x = 4$
- Horizontal asymptote: $y = 1$
- Inverse function: $f^{-1}(y) = \frac{4y + 4}{y - 1}$
- No intersection points exist (complex roots)
Steps to Solve
- Identify the function and vertical asymptote
The function is given as $$f(x) = \frac{x + 4}{x - 4}$$.
To find vertical asymptotes, set the denominator to zero:
$$x - 4 = 0 \implies x = 4$$
Thus, the vertical asymptote is at $x = 4$.
- Find the horizontal asymptote
To find the horizontal asymptote, consider the degrees of the numerator and denominator. For large $|x|$,
$$f(x) \approx \frac{x}{x} = 1.$$
Thus, the horizontal asymptote is at $y = 1$.
- Find the inverse of the function
To find the inverse, set $y = f(x)$:
$$y = \frac{x + 4}{x - 4}.$$
Cross-multiply to solve for $x$:
$$y(x - 4) = x + 4$$
$$yx - 4y = x + 4$$
Rearranging gives:
$$yx - x = 4y + 4$$
$$x(y - 1) = 4y + 4$$
Thus,
$$x = \frac{4y + 4}{y - 1}.$$
So,
$$f^{-1}(y) = \frac{4y + 4}{y - 1}.$$
- Find intersection points of $f(x)$ and $f^{-1}(x)$
Set $f(x) = f^{-1}(x)$:
$$\frac{x + 4}{x - 4} = \frac{4x + 4}{x - 1}.$$
Cross-multiplying yields:
$$(x + 4)(x - 1) = (4x + 4)(x - 4).$$
Expanding both sides results in:
$$x^2 + 4x - x - 4 = 4x^2 - 16x + 4x + 16.$$
Combine like terms:
$$x^2 + 3x - 4 = 4x^2 - 12x + 16.$$
Rearranging gives:
$$3x^2 - 15x + 20 = 0.$$
- Solve the quadratic equation
Using the quadratic formula where $a = 3$, $b = -15$, and $c = 20$:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 3 \cdot 20}}{2 \cdot 3}.$$
Calculate the discriminant:
$$15^2 - (4)(3)(20) = 225 - 240 = -15.$$
Since the discriminant is negative, there are no real intersection points.
- Find the areas enclosed by the graphs of $f(x)$ and $f^{-1}(x)$
Since there are no intersections, the functions do not enclose any area. However, integral setups can be considered for different bounds if needed.
- Vertical asymptote: $x = 4$
- Horizontal asymptote: $y = 1$
- Inverse function: $f^{-1}(y) = \frac{4y + 4}{y - 1}$
- No intersection points exist (complex roots)
More Information
The subject involves understanding asymptotes, function inverses, and intersections in rational functions. The absence of intersection implies no area is enclosed; thus, integrals aren't required.
Tips
- Forgetting to set the denominator to zero when finding vertical asymptotes.
- Not checking for complex roots when solving quadratics.
AI-generated content may contain errors. Please verify critical information
