Problem 48 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Let \(f(x)=\frac{2}{3} x^{3}+x^{2}-12 x+6\). Find the values of \(x\) for which: a. \(f^{\prime}(x)=-12\) b. \(f^{\prime}(x)=0\) c. \(f^{\prime}(x)=12\)

Step-by-Step Solution

Answer

The values of \(x\) for which \(f'(x)=-12\) are \(x=-1\) and \(x=0\), for \(f'(x)=0\) are \(x=-3\) and \(x=2\), and for \(f'(x)=12\) are \(x=-4\) and \(x=3\).

Step 1: Compute the first derivative of the function\(f(x)\)

To compute the first derivative of the given function, differentiate each term with respect to \(x\) using their respective power rule: \[ f'(x) = \frac{d}{dx} \left(\frac{2}{3}x^3 + x^2 - 12x + 6\right) \] Applying the power rule for each term: \[ f'(x) = \frac{2}{3}\cdot3x^{3-1} + 2x^{2-1} - 12x^{1-1} \] Simplify the expression: \[ f'(x) = 2x^2 + 2x - 12 \] Now we have the first derivative, \(f'(x) = 2x^2 + 2x - 12\). Next, we will solve the equations given in parts a, b, and c.

Step 2a: Solve \(f'(x) = -12\) for \(x\)

Substitute -12 for \(f'(x)\) in the equation and solve for \(x\): \[ 2x^2 + 2x -12 = -12 \] Now, simplify the equation: \[ 2x^2 + 2x = 0 \] Factor out the common factor of 2: \[ 2(x^2 + x) = 0 \] Since the product of 2 and the expression in parentheses equals 0, either 2 or the expression in parentheses must equal 0. Since 2 is a constant and cannot be 0, we know that the expression in parentheses must equal 0: \[ x^2 + x = 0 \] To solve this equation, we can factor out an \(x\): \[ x(x + 1) = 0 \] The product of two factors equals 0 if and only if at least one of the factors is equal to 0. So, we have the following possibilities: \[ x = 0 \quad \text{or} \quad x + 1 = 0 \] For the second equation: \[ x = -1 \] So the values of \(x\) for which \(f'(x) = -12\) are \(x=-1\) and \(x=0\).

Step 2b: Solve \(f'(x) = 0\) for \(x\)

Now we need to solve the equation \(f'(x) = 0\) for \(x\): \[ 2x^2 + 2x - 12 = 0 \] Divide the equation by 2: \[ x^2 + x - 6 = 0 \] This equation can be factored into: \[ (x - 2)(x + 3) = 0 \] Thus, the solutions for \(x\) are: \[ x = 2 \quad \text{or} \quad x = -3 \] So the values of \(x\) for which \(f'(x) = 0\) are \(x=-3\) and \(x=2\).

Step 2c: Solve \(f'(x) = 12\) for \(x\)

Now we need to solve the equation \(f'(x) = 12\) for \(x\): \[ 2x^2 + 2x - 12 = 12 \] Simplify the equation: \[ 2x^2 + 2x - 24 = 0 \] Divide the equation by 2: \[ x^2 + x - 12 = 0 \] This equation can be factored into: \[ (x - 3)(x + 4) = 0 \] Thus, the solutions for \(x\) are: \[ x = 3 \quad \text{or} \quad x = -4 \] So the values of \(x\) for which \(f'(x) = 12\) are \(x=-4\) and \(x=3\).

Key Concepts

Function Derivative

A function derivative represents the rate of change of a function with respect to its independent variable. This is a key concept in calculus. Consider the function \( f(x) \). Its derivative, noted as \( f'(x) \), shows how \( f(x) \) changes as \( x \) changes.

To find \( f'(x) \), we differentiate \( f(x) \) using rules of calculus. Differentiation provides us a new function \( f'(x) \). Calculating this means applying derivative rules, such as the power rule, to each term of the function.

This derivative function tells us where the original function is increasing, decreasing, or stationary. It's useful in solving problems related to motion, optimization, and many other fields.

Polynomial Functions

Polynomial functions are a type of mathematical expression made up of terms, each consisting of a coefficient multiplied by a variable raised to a non-negative integer power.

The function \( f(x)=\frac{2}{3} x^{3}+x^{2}-12 x+6 \) is an example of a polynomial function. It involves terms with powers (in this case, 3, 2, and 1), constants, and variables.

Each term's degree is the power of the variable. The degree of the entire polynomial is the highest degree of all its terms. This particular polynomial is of degree 3.

Polynomial functions are continuous and smooth, making them easy to differentiate and integrate, which is highly valuable in calculus.

Solution of Equations

Solving equations involves finding the values of variables that make the equation true. For the given problems in the exercise, we had to find \( x \) values where the derivative equals specific numbers like \(-12\), \(0\), and \(12\).

To solve such problems:

• Set the derivative \( f'(x) \) equal to the desired value.
• Rearrange the equation to move everything to one side, resulting in a standard form equation.
• Factor or use algebraic methods to find values of \( x \).

In these exercises, we often used factoring since it simplifies finding solutions by breaking down the equation into simpler components.

This approach is part of solving algebraic equations and is vital for finding critical points in calculus.

Power Rule in Differentiation

The power rule is a fundamental technique used in differentiation. It helps you find derivatives of polynomial terms efficiently.

To apply the power rule:

• Take any term \( ax^n \), where \( a \) is a coefficient and \( n \) is a positive integer.
• Multiply the term by the exponent \( n \).
• Then, subtract one from the exponent; the result is the new power of \( x \).

For instance, in \( \frac{2}{3}x^3 \), applying the power rule gives \( 3 \cdot \frac{2}{3}x^{3-1}=2x^{2} \). The power rule is simple and quick, making it one of the first rules learned in calculus.

By mastering this rule, you can differentiate a wide range of polynomial functions effortlessly.