Question

In Exercises \(19-30\) , find the extreme values of the function and where they occur. $$y=\frac{3}{2} x^{4}+4 x^{3}-9 x^{2}+10$$

Short answer

The function \(y=\frac{3}{2} x^{4}+4 x^{3}-9 x^{2}+10\) has a maximum at \(x=-1\) with a value of \(y=12.5\) and a minimum at \(x=1\) with a value of \(y=8.5\).

Step-by-step solution

Compute Derivative

Calculate the first derivative (i.e., the slope) of the function \(y=\frac{3}{2} x^{4}+4 x^{3}-9 x^{2}+10\), denoted as \(y'\). The result is \(y'=6x^{3}+12x^{2}-18x\).

Find Critical Points

Identify the critical points by setting \(y'\) equal to 0 and solve for \(x\). After factoring, we get \(y'=6x(x-1)(x+1)\). Solving \(y'=0\) gives us \(x=0, x=-1, x=1\).

Determine Extreme Values

Substitute these critical points into the original function to find the corresponding \(y\) values, which represent the extreme values. We have that when \(x=0\), \(y=10\); when \(x=-1\), \(y=12.5\); and when \(x=1\), \(y=8.5\). We can thus conclude our function has a minimum at \(x=1\) and a maximum at \(x=-1\) (due to the function continuing to rise and fall based on the changing x).

Key concepts

Critical Points

In calculus, critical points are the x-values where the function's first derivative (the slope) is either zero or undefined. These points are essential as they are potential candidates where the function can have extreme values – the highest or lowest points on a graph.

To find critical points, we first calculate the derivative of the function. In our exercise, we found the derivative of the function to be:\(y' = 6x^{3} + 12x^{2} - 18x\).

The next step is to set this derivative equal to zero, since locations where the slope is zero may indicate a peak or valley. After factoring the derivative, we get \(y' = 6x(x - 1)(x + 1)\). Setting \(y' = 0\), we solve for \(x\) and find the critical points: \(x = 0\), \(x = -1\), and \(x = 1\).

By identifying the critical points, we've set the stage to determine if these points correspond to maxima or minima of our function, which is critical for understanding its overall behavior.

First Derivative Test

The First Derivative Test is a powerful tool used to determine whether a critical point is a local maximum, local minimum, or neither. This test involves analyzing the sign (positive or negative) of the first derivative before and after each critical point.

Here's how to apply it: First, list the critical points and then check the sign of the derivative for intervals around these points. A sign change from positive to negative as you pass through a critical point indicates a local maximum. Conversely, a sign change from negative to positive indicates a local minimum. If there’s no sign change, the critical point is neither a maximum nor minimum.

In our exercise, the critical points are \(x = -1\), \(x = 0\), and \(x = 1\). By plugging in values from intervals around these points into the derivative, we would observe the behavior of the slope and apply the First Derivative Test to determine the nature of each critical point.

Maxima and Minima

The terms maxima and minima refer to the highest and lowest points of a function, respectively. Specifically, a maximum is a point where the function's value is larger than at any other point nearby, and a minimum is where the function's value is smaller.

These points can be local (or relative), meaning they are the extreme values within a specific interval or global (or absolute), meaning they are the extreme values in the entire set of the function's domain.

After determining the critical points and using the First Derivative Test, we substitute these critical points back into the original function to find the extreme values. For our function \(y = \frac{3}{2} x^{4} + 4 x^{3} - 9 x^{2} + 10\), the extreme values found were \(y = 12.5\) for \(x = -1\), indicating a maximum, and \(y = 8.5\) for \(x = 1\), indicating a minimum. The value \(y = 10\) for \(x = 0\) suggests a local minimum or maximum based on the typical behavior around such points. By analyzing the changes in the values of the function at these critical points, we've identified the function's maxima and minima.