Question

Finding extrema. For the function \(V(x)=\frac{1}{3} x^{3}+\frac{5}{2} x^{2}-24 x\) (a) Where is the maximum? (b) Where is the minimum?

Short answer

Maximum at \( x = -8 \), minimum at \( x = 3 \).

Step-by-step solution

- Find the First Derivative

Calculate the first derivative of the function \( V(x) \) to find the critical points. The first derivative is given by:\[ V'(x) = \frac{d}{dx}\bigg(\frac{1}{3} x^{3} + \frac{5}{2} x^{2} - 24 x\bigg) \]Applying the differentiation rules, we get:\[ V'(x) = x^{2} + 5x - 24 \]

- Set the First Derivative to Zero

Set the first derivative equal to zero to find the critical points.\[ x^{2} + 5x - 24 = 0 \]This is a quadratic equation that needs to be solved for \( x \).

- Solve the Quadratic Equation

Solve the quadratic equation \( x^{2} + 5x - 24 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 5 \), and \( c = -24 \).Substitute these values into the quadratic formula:\[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} \]\[ x = \frac{-5 \pm \sqrt{25 + 96}}{2} \]\[ x = \frac{-5 \pm \sqrt{121}}{2} \]\[ x = \frac{-5 \pm 11}{2} \]Thus, the solutions are:\[ x = 3 \quad \text{and} \quad x = -8 \]

- Second Derivative Test

Use the second derivative test to determine whether each critical point is a maximum or a minimum. Find the second derivative \( V''(x) \):\[ V''(x) = \frac{d}{dx}(x^{2} + 5x - 24) \]\[ V''(x) = 2x + 5 \]Evaluate \( V''(x) \) at each critical point:For \( x = 3 \):\[ V''(3) = 2(3) + 5 = 11 \]Since \( 11 > 0 \), \( x = 3 \) is a local minimum.For \( x = -8 \):\[ V''(-8) = 2(-8) + 5 = -11 \]Since \( -11 < 0 \), \( x = -8 \) is a local maximum.

Key concepts

First Derivative

The first derivative of a function helps us find critical points, where the slope of the function is zero.
These critical points could be potential maxima or minima.
To find the first derivative, apply differentiation rules to each term of the function.
For example, given the function: \( V(x)=\frac{1}{3} x^{3}+\frac{5}{2} x^{2}-24 x \)
The first derivative is calculated by differentiating each term separately: \(\frac{d}{dx}\bigg(\frac{1}{3} x^{3} + \frac{5}{2} x^{2} - 24 x\bigg) \rightarrow V'(x) = x^{2} + 5x - 24 \)
Setting the first derivative equal to zero helps locate the critical points. Specifically, \( x^{2} + 5x - 24 = 0 \)
These points will be the candidates for either maxima, minima, or points of inflection.

Quadratic Equation

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \).
To solve a quadratic equation like \( x^2 + 5x - 24 = 0 \), we use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For the equation \( x^2 + 5x - 24 = 0 \), the coefficients are \( a = 1 \), \( b = 5 \), and \( c = -24 \).
Plugging these into the formula gives:
\( x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-24)}}{2 \cdot 1} \)
Simplifying the expression under the square root and solving, we find:
\( x = \frac{-5 \pm \sqrt{121}}{2} = \frac{-5 \pm 11}{2} \)
This results in two solutions: \( x = 3 \) and \( x = -8 \).
These are the critical points we will use in the next step.

Second Derivative Test

The second derivative test helps determine if a critical point is a local maximum or minimum.
By taking the second derivative of the original function and evaluating it at the critical points, we can classify them.
For the function: \( V(x)=\frac{1}{3} x^{3}+\frac{5}{2} x^{2}-24 x \), the second derivative is: \(V''(x) = \frac{d}{dx}(x^{2} + 5x - 24) = 2x + 5 \)
Evaluating \(V''(x)\) at the critical points:\( x = 3 \):\( V''(3) = 2(3) + 5 = 11 \) (greater than zero, indicating a local minimum)
\( x = -8 \):\( V''(-8) = 2(-8) + 5 = -11 \) (less than zero, indicating a local maximum).
Thus, \( x = 3 \) is a local minimum, and \( x = -8 \) is a local maximum.