Problem 38 - Thomas Calculus

Question

In Exercises \(35-44,\) find the extreme values of the function and where they occur. $$ y=x^{3}-3 x^{2}+3 x-2 $$

Step-by-Step Solution

Answer

The extreme value is at \(x = 1\), with \(y = -1\).

Step 1: Find the derivative

To find the extreme values of the function, we first need to find the derivative of the function. The function is given as \(y = x^3 - 3x^2 + 3x - 2\). Differentiating term by term, we get: \[\frac{dy}{dx} = 3x^2 - 6x + 3\]

Step 2: Set the derivative to zero

The extreme values occur where the derivative equals zero. Set the derivative \(3x^2 - 6x + 3\) to zero and solve for \(x\). \[3x^2 - 6x + 3 = 0\]

Step 3: Solve the quadratic equation

To solve \(3x^2 - 6x + 3 = 0\), we can factor it. Factoring gives: \[3(x^2 - 2x + 1) = 0 \x^2 - 2x + 1 = (x-1)^2 = 0 \x = 1\] This tells us that \(x = 1\) is the point where a potential extreme value occurs.

Step 4: Determine the nature of the critical point

To determine if the critical point at \(x = 1\) is a maximum or minimum, examine the second derivative. Compute the second derivative of the function:\[\frac{d^2y}{dx^2} = 6x - 6\]Evaluate \(\frac{d^2y}{dx^2}\) at \(x = 1\): \[\frac{d^2y}{dx^2}\bigg|_{x=1} = 6(1) - 6 = 0\]Since the second derivative is zero, this test is inconclusive, indicating a possible inflection point.

Step 5: Verify extreme value by original function value

Since the second derivative test was inconclusive, verify the nature of the function by substituting \(x = 1\) back into the original function to find the function value:\[y(1) = (1)^3 - 3(1)^2 + 3(1) - 2 = -1\] This gives us the point \((1, -1)\). Check values close to \(x=1\) if necessary to confirm nature.

Key Concepts

Derivatives

When working with functions to find extreme values, the first step is to determine the derivative of the given function. In simple terms, a derivative tells us how a function is changing at any given point. It reflects the slope or the rate of change of the function's graph. For example, given the function \(y = x^3 - 3x^2 + 3x - 2\), we find the derivative by differentiating each term:

• The derivative of \(x^3\) is \(3x^2\).
• The derivative of \(-3x^2\) is \(-6x\).
• The derivative of \(3x\) is \(3\).
• The constant \(-2\) has a derivative of \(0\).

Putting it all together, the derivative of the function is \(\frac{dy}{dx} = 3x^2 - 6x + 3\). This derivative helps us find the critical points where the function may have extreme values, such as maxima or minima.

Critical Points

After finding the derivative, the next step is to identify the critical points. Critical points are the locations on the graph of a function where the derivative equals zero or fails to exist. These points are crucial because they are potential candidates for extreme values. For our function, we examine where the derivative \(3x^2 - 6x + 3\) equals zero:

• Set the derivative to zero: \(3x^2 - 6x + 3 = 0\).
• Factor or use algebraic methods to solve for \(x\): \(3(x^2 - 2x + 1) = 0\), which simplifies to \((x-1)^2 = 0\).
• Solving \((x-1)^2 = 0\) gives us \(x = 1\) as the critical point.

At \(x = 1\), the derivative is zero, indicating a possible extreme value. Therefore, at \(x = 1\), the function potentially changes from increasing to decreasing or vice versa.

Second Derivative Test

To further determine the nature of a critical point, we use the second derivative test. This test tells us whether a critical point is a local maximum, minimum, or an inflection point by examining the concavity of the function at that point. The second derivative of our function is the derivative of the first derivative:

• The first derivative is \(3x^2 - 6x + 3\).
• The second derivative is \(\frac{d^2y}{dx^2} = 6x - 6\).

Let's evaluate the second derivative at the critical point \(x = 1\):

• \(\frac{d^2y}{dx^2}\bigg|_{x=1} = 6(1) - 6 = 0\).

The result is zero, which means that the test is inconclusive for determining a maximum or minimum. This suggests that there might be an inflection point. To confirm, we can check the original function value or test nearby values to see how the function behaves around \(x = 1\). In our case, substitution gives the point \((1, -1)\). This point does not provide a clear extreme value, hence further investigation or methods might be needed to gain more insight.