Question

A function \(f(x)\) is given. Find the critical points of \(f\) and use the Second Derivative Test, when possible, to determine the relative extrema. \(f(x)=2 x^{3}-3 x^{2}+9 x+5\)

Short answer

No real critical points exist; hence, no relative extrema via the Second Derivative Test.

Step-by-step solution

Find the Derivative of the Function

To find critical points, first determine the derivative of the function, since critical points occur where this derivative equals zero or is undefined. The derivative of the given function is\[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 + 9x + 5) = 6x^2 - 6x + 9. \]

Solve for Critical Points

Set the derivative equal to zero to find the critical points:\[ 6x^2 - 6x + 9 = 0. \]Solving this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 6, b = -6, c = 9 \):\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 6 \cdot 9}}{2 \cdot 6}. \]The discriminant \((-6)^2 - 4 \cdot 6 \cdot 9 = 36 - 216 = -180,\) which is negative. Since the discriminant is negative, there are no real solutions, thus no real critical points.

Determine Relative Extrema Using the Second Derivative Test

Since no real critical points exist, a Second Derivative Test can't be applied. If there were real critical points, you would substitute them into the second derivative \( f''(x) \) to determine concavity. However, this step is not feasible here because the function has no real critical points.

Key concepts

Second Derivative Test

The Second Derivative Test is a valuable tool in calculus, used to determine the nature of critical points found on the graph of a function. This test involves two main steps: first, finding critical points by setting the first derivative equal to zero, and second, using the second derivative to test the concavity of the function at those points.
To apply the Second Derivative Test, follow these steps:

• Calculate the second derivative of the function, denoted as \( f''(x) \).
• Evaluate \( f''(x) \) at each critical point.

If \( f''(x) > 0 \) at a critical point, the function is concave up, indicating a local minimum. Conversely, if \( f''(x) < 0 \), the function is concave down, suggesting a local maximum. If \( f''(x) = 0 \), the test is inconclusive, and the nature of the critical point might not be determined using this method alone.
It's essential only to use the Second Derivative Test where real critical points exist. If, as in this case, critical points do not have real values, the test cannot be applied, which means determining the relative extrema through this method is not feasible.

Relative Extrema

Relative extrema refer to the highest or lowest points in a particular region of a graph, known as local maxima or minima. Unlike absolute extrema, which are the extreme values over the entire range of a function, relative extrema are restricted to a neighborhood or interval.
To find relative extrema, you first need to determine the critical points of the function, where the first derivative \( f'(x) \) is zero or undefined. These points are potential candidates for local maxima or minima.
Once critical points are established, you can use the Second Derivative Test to assess whether these points are maxima, minima, or saddle points. This involves checking the concavity of the function at each critical point using the second derivative.
However, if there are no real critical points, as is the case here with a negative discriminant, there are no relative extrema to be found. Therefore, interpreting the relative extrema relies on identifying valid real solutions for the critical points.

Polynomial Functions

Polynomial functions are mathematical expressions that involve sums of powers of a variable, usually denoted as \( x \). They take the form \( a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \), where \( n \) is a non-negative integer and \( a_n, a_{n-1}, \ldots, a_0 \) are coefficients.
These functions are continuous and smooth, making them useful in various applications, from physics to economics. Derived from their general form, polynomial functions have derivatives and critical points that are solvable and analyzable, much like the function \( f(x) = 2x^3 - 3x^2 + 9x + 5 \) in the exercise.
When dealing with critical points in polynomial functions, setting the derivative to zero allows us to solve for values of \( x \) where the slope is zero, indicating potential extrema. If the derivative yields a quadratic equation, like in the example, the discriminant helps determine the nature of the solutions.
Polynomial functions exhibit predictable behavior, due to their defined degree \( n \). For instance, a cubic polynomial like \( 2x^3 \) has at most two turning points, corresponding to critical points, which may or may not have real solutions depending on the coefficients involved.