Question

Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible ) whether they correspond to local maxima or local minima. $$f(x)=x^{3}-13 x^{2}-9 x$$

Short answer

Question: Determine the nature of the critical points of the function $$f(x) = x^3 - 13x^2 - 9x$$. Answer: The function has a local maximum at $$x_1 = 1$$ and a local minimum at $$x_2 = 9$$.

Step-by-step solution

1. Finding the first derivative

To find the first derivative of the function, we'll use the power rule. The first derivative, denoted as $$f'(x)$$, is: $$f'(x) = 3x^2 - 26x - 9$$

2. Determine the critical points

Critical points occur where the derivative is zero or undefined. In our case, the derivative is a polynomial and hence always defined. To find the critical points, we set $$f'(x)$$ equal to zero and solve for $$x$$: $$0 = 3x^2 - 26x - 9$$ We are dealing with a quadratic equation. To solve it, we can use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Where $$a = 3$$, $$b = -26$$, and $$c = -9$$. Then: $$x = \frac{26 \pm \sqrt{(-26)^2 - 4(3)(-9)}}{6}$$ $$x = \frac{26 \pm \sqrt{676 + 108}}{6}$$ $$x = \frac{26 \pm \sqrt{784}}{6}$$ $$x = \frac{26 \pm 28}{6}$$ So we have two critical points: $$x_1 = 1$$ and $$x_2 = 9$$.

3. Find the second derivative

Next, we need to find the second derivative of the function, denoted as $$f''(x)$$. It is the derivative of the first derivative: $$f''(x) = 6x - 26$$

4. Apply the Second Derivative Test

Now we apply the Second Derivative Test, which states that if $$f''(x)$$ is positive at a critical point, then the function has a local minimum at that point. If $$f''(x)$$ is negative, the function has a local maximum at that point. Evaluate $$f''(x_1 = 1)$$: $$f''(1) = 6(1) - 26 = -20$$ Since $$f''(1)$$ is negative, the function has a local maximum at $$x_1 = 1$$. Evaluate $$f''(x_2 = 9)$$: $$f''(9) = 6(9) - 26 = 28$$ Since $$f''(9)$$ is positive, the function has a local minimum at $$x_2 = 9$$. In conclusion, the function $$f(x) = x^3 - 13x^2 - 9x$$ has a local maximum at $$x_1 = 1$$ and a local minimum at $$x_2 = 9$$.

Key concepts

First Derivative

To find the critical points of a function, the first step is to calculate the first derivative. The first derivative, denoted as \( f'(x) \), provides us with the slope of the tangent line to the curve at any point \( x \). It's a measure of how the function value changes with a small change in \( x \).
In the equation \( f(x) = x^3 - 13x^2 - 9x \), we apply the power rule of differentiation. The power rule states that for any term \( ax^n \), its derivative is \( n \cdot ax^{n-1} \). Thus, for our function:

• The derivative of \( x^3 \) is \( 3x^2 \).
• The derivative of \( -13x^2 \) is \( -26x \).
• The derivative of \( -9x \) is \( -9 \).

Piecing these together, we get \( f'(x) = 3x^2 - 26x - 9 \).
This equation will help us find where the slope is zero or undefined, indicating potential critical points.

Second Derivative Test

Once the critical points are identified using the first derivative, the next step is to determine the nature of these points, which could be a local maximum, minimum, or a saddle point. We use the Second Derivative Test for this purpose.
Calculate the second derivative, \( f''(x) \), from the first derivative \( f'(x) = 3x^2 - 26x - 9 \). Differentiating once more, applying the power rule, we get \( f''(x) = 6x - 26 \).
The Second Derivative Test works as follows:

• If \( f''(x) > 0 \) at a critical point, the function is concave upwards there, indicating a local minimum.
• If \( f''(x) < 0 \) at a critical point, the function is concave downwards there, indicating a local maximum.
• If \( f''(x) = 0 \), the test is inconclusive, and other methods must be used to determine the point's nature.

By evaluating \( f''(1) \) and \( f''(9) \), we concluded:
- \( f''(1) = -20 \) (local maximum).
- \( f''(9) = 28 \) (local minimum).

Quadratic Formula

Quadratic equations often arise when determining critical points from the first derivative. In our case, the critical points were found by solving \( 3x^2 - 26x - 9 = 0 \).
The quadratic formula is a reliable and powerful tool used to solve any quadratic equation \( ax^2 + bx + c = 0 \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]For our specific equation, the coefficients are:

• \( a = 3 \)
• \( b = -26 \)
• \( c = -9 \)

Plugging these into the formula:\[ x = \frac{26 \pm \sqrt{676 + 108}}{6} \]\[ x = \frac{26 \pm \sqrt{784}}{6} \]\[ x = \frac{26 \pm 28}{6} \]This results in two solutions: \( x_1 = 1 \) and \( x_2 = 9 \), which are the critical points. The quadratic formula thus simplifies solving such equations efficiently.

Local Maximum and Minimum

Understanding local maxima and minima is key in analyzing the behavior of functions. A local maximum is a point where the function reaches a peak, whereas a local minimum is a point where the function reaches a trough. These points illustrate where a function changes direction in its increasing or decreasing behavior.
In our exercise, after finding the critical points \( x_1 = 1 \) and \( x_2 = 9 \), and employing the Second Derivative Test:

• At \( x = 1 \), \( f''(1) = -20 \), indicating a concave downwards curve at this point. Therefore, \( x = 1 \) is a local maximum.
• At \( x = 9 \), \( f''(9) = 28 \), indicating a concave upwards curve at this point. Hence, \( x = 9 \) is a local minimum.

Local extrema provide valuable insights into how a function behaves over an interval, helping identify turning points in its graph.