Problem 60 - University Calculus: Early Transcendentals

Question

Find the extreme values (absolute and local) of the function over its natural domain, and where they occur. $$y=\sqrt{3+2 x-x^{2}}$$

Step-by-Step Solution

Answer

The absolute maximum is 2 at x=1; absolute minima are 0 at x=-1 and x=3.

Step 1: Determine the Domain

The function provided is \( y = \sqrt{3 + 2x - x^2} \). Since this is a square root function, the expression inside the square root must be non-negative. Therefore, we need to solve \( 3 + 2x - x^2 \geq 0 \). This is a quadratic inequality, which can be solved by finding the roots of the corresponding equation \( 3 + 2x - x^2 = 0 \).

Step 2: Solve the Quadratic Equation

Rewrite the quadratic equation in standard form: \(-x^2 + 2x + 3 = 0\). To find the roots, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = -1 \), \( b = 2 \), and \( c = 3 \). Thus, \( x = \frac{-2 \pm \sqrt{2^2 - 4(-1)(3)}}{2(-1)} \), which simplifies to \( x = \frac{-2 \pm \sqrt{4 + 12}}{-2} = \frac{-2 \pm 4}{-2} \). The roots are \( x = -1 \) and \( x = 3 \).

Step 3: Identify the Domain Interval

Since the quadratic \( 3 + 2x - x^2 \) opens downwards (as the coefficient of \( x^2 \) is negative), the values between \( x = -1 \) and \( x = 3 \) make the quadratic non-negative. Therefore, the domain of the function is \( x \in [-1, 3] \).

Step 4: Find Critical Points

To find critical points, compute the derivative of the function. The derivative of \( y = \sqrt{3 + 2x - x^2} \) is \( y' = \frac{1}{2\sqrt{3+2x-x^2}} (2-2x) \). Setting \( y' = 0 \) gives \( 2 - 2x = 0 \), which implies \( x = 1 \). The critical point is \( x = 1 \).

Step 5: Evaluate Function at Critical Points and Endpoints

The extreme values can occur at critical points or endpoints of the domain. Evaluate \( y \) at \( x = -1 \), \( x = 1 \), and \( x = 3 \): - \( y(-1) = \sqrt{3 + 2(-1) - (-1)^2} = \sqrt{0} = 0 \)- \( y(1) = \sqrt{3 + 2(1) - (1)^2} = \sqrt{4} = 2 \)- \( y(3) = \sqrt{3 + 2(3) - (3)^2} = \sqrt{0} = 0 \)

Step 6: Determine Extreme Values

Compare the function values obtained at the critical point and endpoints. The absolute maximum value of \( y \) is 2 at \( x = 1 \). The absolute minimum values of \( y \) are 0 at \( x = -1 \) and \( x = 3 \).

Key Concepts

Quadratic Inequality

Solving quadratic inequalities is a fundamental step when identifying domains for functions involving square roots. In this problem, the inequality \(3 + 2x - x^2 \geq 0\) ensures that the expression under the square root remains non-negative. This is crucial because the square root of a negative number is not defined in real numbers. To solve such inequalities:

• Convert the inequality into an equation: \(3 + 2x - x^2 = 0\).
• Find the roots using the quadratic formula, where \( a = -1 \), \( b = 2 \), and \( c = 3 \).
• Calculate: \( x = \frac{-2 \pm \sqrt{4 + 12}}{-2} \), resulting in roots \( x = -1 \) and \( x = 3 \).
• Determine the intervals. A downward-opening parabola, affected by the negative \( x^2 \) term, means the inequality holds within the roots, hence the domain \( x \in [-1, 3] \).

This solution determines where the function is real and provides critical bounds within which further exploration occurs.

Critical Points

Critical points are where a function's rate of change, indicated by its derivative, is zero or does not exist. These are potential locations for local extrema (maxima or minima). To find a critical point:

• Differentiate the function. The derivative of \( y = \sqrt{3 + 2x - x^2} \) is \( y' = \frac{1}{2\sqrt{3+2x-x^2}} (2-2x) \).
• Set the derivative equal to zero: \( 2 - 2x = 0 \) implies \( x = 1 \).
• Consider where the derivative doesn't exist; however, this typically occurs outside the domain \( x \in [-1, 3] \).

These critical points must be checked along with endpoint values to determine actual extrema. Here, \( x = 1 \) is a candidate that must be evaluated for its function value.

Derivative of Functions

The derivative of a function describes its instantaneous rate of change and is crucial in finding critical points and behavior. For functions involving square roots, chain rule application is common. In this case:

• The function \( y = \sqrt{3 + 2x - x^2} \) yields a derivative through the chain rule: \( y' = \frac{1}{2\sqrt{3 + 2x - x^2}} \, (2 - 2x) \).
• This derivative helps identify critical points, where changes in the function's behavior occur.
• Calculating derivatives precisely aids in understanding where the function increases, decreases, or maintains constant values within a domain.

Using derivatives, one can explore not just if extreme values occur, but also how the function behaves in different intervals.