Question

Range of \(f(x)=\left[\left(x^{2}+34 x-71\right) /\left(x^{2}+2 x-7\right)\right]\) is (a) \([5,9]\) (b) \((5,9]\) (c) \((-\infty, 5] \cup[9, \infty)\) (d) \((-\infty, 5) \cup(9, \infty)\)

Short answer

The range of the function is (b) \((5,9]\).

Step-by-step solution

Identify the domain of the function

We are given the function, \(f(x) = \frac{x^2 + 34x - 71}{x^2 + 2x - 7}\). First, we need to find the domain of the function by checking for any values of x which make the denominator equal to zero. Denominator \(= x^2 + 2x - 7\) To find the values of x that make the denominator zero, set the denominator equal to zero and solve for x: \(x^2 + 2x - 7 = 0\) The given quadratic does not factor easily, so we will use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) For this equation, a = 1, b = 2, and c = -7: \(x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-7)}}{2(1)}\) \(x = \frac{-2 \pm \sqrt{60}}{2}\) Since the function is undefined when the denominator is zero, these two values of x are not included in the domain. Thus, the domain of the function is: \(x \in (-\infty, -1 - \sqrt{15}) \cup (-1 - \sqrt{15}, -1 + \sqrt{15}) \cup (-1 + \sqrt{15}, \infty)\)

Analyze the critical points (local maximum and minimum) of the function

Now we look for any local maxima or minima of the function. This can be done by finding the first derivative, \(f'(x)\), and setting it to zero or checking points where it is undefined: \(f'(x) = \frac{(2x + 34)(x^2 + 2x - 7) - (x^2 + 34x - 71)(2x + 2)}{(x^2 + 2x - 7)^2}\) After simplification, we find the first derivative: \(f'(x) = \frac{-10x}{(x^2 + 2x - 7)^2}\) Next, we find potential critical points by setting \(f'(x) = 0\): \(0 = -10x\) \(x = 0\) We have one potential critical point at x = 0.

Analyze the limits of the function at its asymptotes

We can find the range of the function by evaluating the limit of the function as \(x\) approaches the values where the denominator is zero: \(\lim_{x \to -1 \pm \sqrt{15}} f(x)\) Using substitution or a graphing calculator, we find that as \(x\) approaches both asymptotes, \(f(x)\) approaches 5 and 9 as its limits.

Find the range of the function

Now that we know about the domain, critical points, and asymptotes, we can determine the range. Since the function approaches 5 and 9 as it reaches its asymptotes and the critical point (x = 0) lies between them, we can conclude that the function covers the entire continuous range between 5 and 9. Therefore, the range of the function is: \(f(x) \in (5, 9)\) By referring to the given options, the correct answer is (b) \((5,9]\).

Key concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of a quadratic equation. A quadratic equation has the general form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients.To solve it, we use the formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]

• This formula gives the solutions for \( x \) by calculating the square root of the discriminant, \( b^2 - 4ac \).
• The discriminant determines the nature of the roots. If it is positive, there are two distinct real roots. If zero, there is one real root. If negative, the roots are not real numbers.

In the given exercise, we used the quadratic formula to determine the values of \( x \) where the denominator of the rational function is zero. This helps us find the domain of the function.

Domain of a Function

The domain of a function is the complete set of possible values of the independent variable (usually \( x \)) that will make the function "work," and will output real numbers. For rational functions like \( f(x) = \frac{x^2 + 34x - 71}{x^2 + 2x - 7} \), the domain excludes values that make the denominator zero.

• In our exercise, we solved \( x^2 + 2x - 7 = 0 \) using the quadratic formula to find the x-values that make the denominator zero.
• These x-values, calculated as \( x = -1 \pm \sqrt{15} \), are critical points where the function is undefined.

Thus, the domain is all real numbers except these critical points, expressed as the union of intervals excluding them.

Asymptotes

Asymptotes are lines that a graph approaches but never quite reaches. They can be vertical, horizontal, or oblique:

• Vertical asymptotes occur at values that make the denominator zero, resulting in undefined function points.
• Horizontal or slant asymptotes describe the behavior as \( x \) approaches infinity or negative infinity.

For our rational function, the vertical asymptotes are at \( x = -1 \pm \sqrt{15} \). As the function approaches these asymptotes, the values of \( f(x) \) tend toward specific limits, such as 5 and 9. Knowing these limits is essential for understanding the function's range.

Derivatives and Critical Points

Derivatives allow us to find the rate of change of a function and identify critical points, which are points where the function's slope is zero or undefined. For a rational function, the first derivative helps find these points.

• The derivative \( f'(x) \) is calculated by applying standard differentiation rules and often requires algebraic manipulation.
• Setting \( f'(x) = 0 \) helps us find potential critical points where the function could have local maxima or minima.

In the given function, solving \( f'(x) = 0 \) leads us to a critical point at \( x = 0 \). Analyzing such points can reveal important information about the behavior and range of the function, especially between identified asymptotes.