Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 5-26, identify the critical points and find the maximum value and minimum value on the given interval. $$ \Psi(x)=x^{2}+3 x ; I=[-2,1] $$

Step-by-Step Solution

Answer

Maximum value: 4 at \( x = 1 \); Minimum value: \(-2.25\) at \( x = -\frac{3}{2} \).

Step 1: Find the derivative

To find the critical points, we first need to find the derivative of \( \Psi(x) \). The function is \( \Psi(x) = x^2 + 3x \). Using basic derivative rules:\[ \Psi'(x) = 2x + 3 \]

Step 2: Set the derivative to zero

Next, set \( \Psi'(x) = 0 \) to find the critical points:\[ 2x + 3 = 0 \]Solve for \( x \):\[ x = -\frac{3}{2} \]

Step 3: Evaluate the function at critical points and endpoints

The interval given is \([-2, 1]\). We need to evaluate \( \Psi(x) \) at \( x = -\frac{3}{2} \), as well as the endpoints \( x = -2 \) and \( x = 1 \).1. \( \Psi(-2) = (-2)^2 + 3(-2) = 4 - 6 = -2 \)2. \( \Psi(-\frac{3}{2}) = \left( -\frac{3}{2} \right)^2 + 3\left(-\frac{3}{2}\right) = \frac{9}{4} - \frac{9}{2} = \frac{9}{4} - \frac{18}{4} = -\frac{9}{4} \)3. \( \Psi(1) = 1^2 + 3(1) = 1 + 3 = 4 \)

Step 4: Identify maximum and minimum values

Compare the values computed:- \( \Psi(-2) = -2 \)- \( \Psi(-\frac{3}{2}) = -\frac{9}{4} \approx -2.25 \)- \( \Psi(1) = 4 \)The maximum value within the interval is \( 4 \) and occurs at \( x = 1 \). The minimum value is approximately \(-2.25\) and occurs at \( x = -\frac{3}{2} \).

Key Concepts

Derivative

The derivative is a crucial concept in calculus that helps us understand how a function changes. It essentially shows the rate of change or the slope of the function at any given point. To find the derivative of a function, we apply certain rules of differentiation. In the provided exercise, the function is \( \Psi(x) = x^2 + 3x \). To find its derivative, we apply the power rule. The derivative of \( x^2 \) is \( 2x \), and the derivative of \( 3x \) is 3. Combining these, we get \( \Psi'(x) = 2x + 3 \). This derivative will help us identify where the slope of the function is zero, indicating potential critical points where maximum or minimum values might occur.

Interval

In calculus, an interval is a range of values within which we are analyzing a function. It defines the portion of the graph we're interested in. The interval can be closed, open, or half-open, based on whether it includes the endpoints or not. In this exercise, the interval given is \([-2,1]\). This means we're only considering this section of the \( x \)-axis, and any critical points or extreme values we find should also lie within this range. We evaluate critical points and endpoints within this interval because our solutions are only valid within this specified part of the graph.

Maximum Value

The maximum value of a function on a closed interval is the highest point on the graph within that interval. To find it, we evaluate the function at critical points and endpoints. This includes any point where the derivative equals zero (as these points can represent peaks) and the boundaries of the interval. In our exercise, we calculated the function value at \( x = -2 \), \( x = -\frac{3}{2} \), and \( x = 1 \). After comparing these values, we found that the maximum value is \( 4 \), which occurs at the endpoint \( x = 1 \). Therefore, the graph reaches its highest point within the interval at this location.

Minimum Value

The minimum value of a function on a given interval is the lowest point it reaches. This point can be at a critical point or at the endpoints of the interval. By examining the function values at these key points, we determine where the function dips the lowest. In the solution for the exercise, we evaluated \( \Psi(x) \) at the given points and found that the minimum value of approximately \(-2.25\) occurs at \( x = -\frac{3}{2} \), a critical point where the derivative \( \Psi'(x) = 0 \). This reflects the lowest point of the function on the graph within the specified interval.