Problem 9 - Calculus with Concepts in Calculus

Question

Use the First Derivative Test to determine the relative extreme values (if any) of the function. $$ f(x)=-3 x^{2}+3 x+7 $$

Step-by-Step Solution

Answer

The relative maximum value is \( \frac{31}{4} \) at \( x = \frac{1}{2} \).

Step 1: Find the derivative of the function

To find critical points, we first need to find the derivative of the function. The given function is \( f(x) = -3x^2 + 3x + 7 \). Using the power rule, the derivative \( f'(x) \) is calculated as follows:\[ f'(x) = \frac{d}{dx}(-3x^2) + \frac{d}{dx}(3x) + \frac{d}{dx}(7) \]\[ f'(x) = -6x + 3 \]

Step 2: Find critical points

Critical points occur where the derivative is zero or undefined. Since \( f'(x) = -6x + 3 \), we set it equal to zero to find critical points.\[ -6x + 3 = 0 \]Solving for \( x \), we get:\[ -6x = -3 \]\[ x = \frac{1}{2} \]

Step 3: Use the First Derivative Test

The First Derivative Test involves looking at the sign of \( f'(x) \) around the critical point. We'll test intervals around \( x = \frac{1}{2} \).- Choose \( x = 0 \) (which is less than \( \frac{1}{2} \)): \[ f'(0) = -6(0) + 3 = 3 \] The sign is positive.- Choose \( x = 1 \) (which is greater than \( \frac{1}{2} \)): \[ f'(1) = -6(1) + 3 = -3 \] The sign is negative.Since \( f'(x) \) changes from positive to negative at \( x = \frac{1}{2} \), there is a relative maximum at this point.

Step 4: Determine the relative extreme value

To find the value of the relative maximum, we substitute the critical point back into the original function.\[ f\left(\frac{1}{2}\right) = -3\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right) + 7 \]Calculate:\[ f\left(\frac{1}{2}\right) = -3\left(\frac{1}{4}\right) + \frac{3}{2} + 7 \]\[ f\left(\frac{1}{2}\right) = -\frac{3}{4} + \frac{3}{2} + 7 \]Combine the terms:\[ f\left(\frac{1}{2}\right) = \frac{3}{2} - \frac{3}{4} + 7 = \frac{3}{4} + 7 = \frac{31}{4} \]

Step 5: Conclusion: Identify the relative extreme value

The function \( f(x) = -3x^2 + 3x + 7 \) has a relative maximum at \( x = \frac{1}{2} \) with the value of \( \frac{31}{4} \).

Key Concepts

Critical Points

Critical points are essential when analyzing the behavior of a function, as they indicate where the function may have a relative maximum, minimum, or point of inflection. These points are found where the derivative of the function equals zero or is undefined. For the function given by \( f(x) = -3x^2 + 3x + 7 \), the derivative is calculated to find these points. The derivative, \( f'(x) = -6x + 3 \), is set equal to zero:

• \( -6x + 3 = 0 \)
• Solving, we get \( x = \frac{1}{2} \)

This calculation shows that there is a critical point at \( x = \frac{1}{2} \). Understanding critical points helps in determining intervals where the function increases or decreases, and is key for identifying relative extrema.

Relative Maximum

A relative maximum is a point where a function changes direction from increasing to decreasing, reaching a peak within a certain interval. To identify a relative maximum using the First Derivative Test, you must analyze the sign changes of the derivative around the critical points. For our function, \( f(x) \), this involves:

• Testing values less than the critical point: For \( x = 0 \), \( f'(0) = 3 \), which is positive, indicating the function is increasing.
• Testing values greater than the critical point: For \( x = 1 \), \( f'(1) = -3 \), which is negative, indicating the function is decreasing.

Since the derivative shifts from positive to negative as \( x \) passes through \( \frac{1}{2} \), the function has a relative maximum at this critical point. The actual value of the function at this point \( x = \frac{1}{2} \) is found by substituting back into the original function, resulting in \( f\left(\frac{1}{2}\right) = \frac{31}{4} \). This means the highest point occurs at \( x = \frac{1}{2} \) with a function value of \( \frac{31}{4} \).

Derivative of a Function

The derivative of a function represents the rate of change of the function's value with respect to its variable; it effectively provides the slope of the tangent line at any point on the function's graph. Calculating the derivative is crucial for finding critical points and determining the behavior of the function. For our quadratic function \( f(x) = -3x^2 + 3x + 7 \), its derivative \( f'(x) = -6x + 3 \) is computed by applying the power rule:

• Power Rule: \( \frac{d}{dx}x^n = nx^{n-1} \)
• For \( -3x^2 \): \( -6x \)
• For \( 3x \): \( 3 \)
• The constant term \( 7 \) has a derivative of 0.

Understanding and calculating derivatives allow us to efficiently find points where the function's behavior changes, which are just what we need to analyze increasing and decreasing intervals, critical points, and to identify maxima and minima. In solving this exercise, mastering derivatives is essential for employing the First Derivative Test accurately.