Problem 30 - Calculus Single Variable

Question

Find the absolute minimum value and absolute maximum value of the given function on the given interval. \(f(x)=\left(x^{2}-3\right) /(x+2) ;[-3 / 2,1]\)

Step-by-Step Solution

Answer

The absolute minimum is -2, and the absolute maximum is -\(\frac{2}{3}\).

Step 1: Understanding the Function and Interval

We are given the function \( f(x) = \frac{x^2 - 3}{x + 2} \) and need to find its absolute minimum and maximum over the interval \([-\frac{3}{2}, 1]\). The first step is to ensure the function is defined on this interval. Note that \( f(x) \) is not continuous at \( x = -2 \) due to the denominator, but \( -2 \) is not in our interval.

Step 2: Find the Critical Points

To find the critical points, differentiate the function with respect to \( x \). The derivative is \( f'(x) = \frac{(2x)(x+2) - (x^2 - 3) \cdot 1}{(x+2)^2} \). Simplify to find the numerator: \( 2x^2 + 4x - x^2 + 3 = x^2 + 4x + 3 \). Set the numerator equal to 0: \( x^2 + 4x + 3 = 0 \). Factor the quadratic: \( (x+1)(x+3) = 0 \). Thus, the critical points are \( x = -1 \) and \( x = -3 \). Only \( x = -1 \) lies in the interval.

Step 3: Evaluate the Function at Critical and End Points

Evaluate \( f(x) \) at the critical point \( x = -1 \) and at the endpoints \( x = -\frac{3}{2} \) and \( x = 1 \):- \( f(-1) = \frac{(-1)^2 - 3}{-1 + 2} = \frac{1 - 3}{1} = -2 \).- \( f(-\frac{3}{2}) = \frac{(-\frac{3}{2})^2 - 3}{-\frac{3}{2} + 2} = \frac{\frac{9}{4} - 3}{\frac{1}{2}} = \frac{-\frac{3}{4}}{\frac{1}{2}} = -\frac{3}{2} \).- \( f(1) = \frac{1 - 3}{1 + 2} = \frac{-2}{3} = -\frac{2}{3} \).

Step 4: Determine the Absolute Minimum and Maximum

The function values calculated in Step 3 are:- \( f(-1) = -2 \), - \( f(-\frac{3}{2}) = -\frac{3}{2} \), - \( f(1) = -\frac{2}{3} \).Now, compare these values to identify the absolute minimum and maximum. The absolute minimum value is \( -2 \), and the absolute maximum value is \( -\frac{2}{3} \).

Key Concepts

Critical Points

Finding critical points is essential in calculus optimization problems. Critical points are where a function's derivative is zero or undefined. These points indicate potential locations for local minima or maxima. To find them for the function \( f(x) = \frac{x^2 - 3}{x + 2} \), we first need to compute its derivative. The derivative \( f'(x) \) is obtained using the quotient rule, which states: if you have \( \frac{u}{v} \), then the derivative is \( \frac{u'v - uv'}{v^2} \). Applying this to our function gives the derivative as \( \frac{(2x)(x+2) - (x^2 - 3) \cdot 1}{(x+2)^2} \).Once simplified, the numerator becomes \( x^2 + 4x + 3 \), leading us to set \( x^2 + 4x + 3 = 0 \) to find critical points. This factors to \( (x+1)(x+3) = 0 \), yielding \( x = -1 \) and \( x = -3 \). However, only \( x = -1 \) is within our interval \([-rac{3}{2}, 1]\), thus it is the only critical point to consider.In summary, critical points offer potential locations for optimization and finding them involves differentiating and solving for zeros in the derivative.

Function Evaluation

After identifying a critical point, evaluating the function at these points along with the interval's endpoints is crucial. This process helps determine which values might be absolute minimums or maximums. For the function \( f(x) = \frac{x^2 - 3}{x + 2} \), we need to check \( f(x) \) at \( x = -1 \) (the critical point), \( x = -\frac{3}{2} \), and \( x = 1 \) (the endpoints of the interval).Evaluating these we find:

• At \( x = -1 \), \( f(-1) = -2 \)
• At \( x = -\frac{3}{2} \), \( f(-\frac{3}{2}) = -\frac{3}{2} \)
• At \( x = 1 \), \( f(1) = -\frac{2}{3} \)

To evaluate efficiently, substitute each point into the function expression and simplify. These evaluations will guide us in comparing function values and pinpointing minima and maxima across the specified interval.Through understanding the values of \( f(x) \) at critical and boundary points, we systematically uncover the function's behavior on the given interval.

Interval Analysis

Interval analysis in calculus involves examining a function over a specific interval to determine its absolute extrema. This process combines information from critical and endpoint evaluations, ensuring we identify absolute minimum and maximum values.For \( f(x) = \frac{x^2 - 3}{x + 2} \) on the interval \([-rac{3}{2}, 1]\), after evaluating at critical and endpoint values, we have:

• \( f(-1) = -2 \)
• \( f(-\frac{3}{2}) = -\frac{3}{2} \)
• \( f(1) = -\frac{2}{3} \)

By comparing these, the lowest value \( -2 \) occurs at \( x = -1 \), indicating the absolute minimum. Similarly, the largest value, \(-\frac{2}{3}\), occurs at \( x = 1 \), signifying the absolute maximum.When we apply interval analysis, we're ensuring that overall behavior along the interval is accounted for, making it possible to definitively state where our function peaks and reaches its lowest point. This method is vital in applications where precise optimization is needed.