Problem 36 - Thomas Calculus

Question

In Exercises \(35-44,\) find the extreme values of the function and where they occur. $$ y=x^{3}-2 x+4 $$

Step-by-Step Solution

Answer

Local minimum at \(x=\sqrt{\frac{2}{3}}\); local maximum at \(x=-\sqrt{\frac{2}{3}}\).

Step 1: Find the First Derivative

To find the critical points where extreme values might occur, we need the first derivative of the function. The function is given by \(y = x^3 - 2x + 4\). The derivative \(y'\) of this function is found by differentiating each term: \[ y' = \frac{d}{dx}(x^3) - \frac{d}{dx}(2x) + \frac{d}{dx}(4) = 3x^2 - 2. \]

Step 2: Set the First Derivative to Zero

We set the first derivative equal to zero to find the critical points: \[ 3x^2 - 2 = 0. \] This will help us find where the function has potential maxima, minima, or points of inflection.

Step 3: Solve for Critical Points

Solve the equation from the previous step for \(x\). \[ 3x^2 - 2 = 0 \] \[ 3x^2 = 2 \] \[ x^2 = \frac{2}{3} \] \[ x = \pm\sqrt{\frac{2}{3}}. \] Thus, the critical points occur at \(x = \sqrt{\frac{2}{3}}\) and \(x = -\sqrt{\frac{2}{3}}\).

Step 4: Find the Second Derivative

The second derivative will help determine the nature of the critical points. Differentiate \(y' = 3x^2 - 2\): \[ y'' = \frac{d}{dx}(3x^2) = 6x. \]

Step 5: Determine Nature of Critical Points

Substitute each critical point into the second derivative.- For \(x = \sqrt{\frac{2}{3}}\): \[ y''(\sqrt{\frac{2}{3}}) = 6 \cdot \sqrt{\frac{2}{3}} > 0, \] indicating a local minimum.- For \(x = -\sqrt{\frac{2}{3}}\): \[ y''(-\sqrt{\frac{2}{3}}) = 6 \cdot (-\sqrt{\frac{2}{3}}) < 0, \] indicating a local maximum.

Step 6: Calculate the Extreme Values

Substitute the critical points into the original function to find the extreme values.- For \(x = \sqrt{\frac{2}{3}}\): \[ y = (\sqrt{\frac{2}{3}})^3 - 2(\sqrt{\frac{2}{3}}) + 4. \]- For \(x = -\sqrt{\frac{2}{3}}\): \[ y = (-\sqrt{\frac{2}{3}})^3 - 2(-\sqrt{\frac{2}{3}}) + 4. \]

Key Concepts

Understanding the First Derivative

The first derivative of a function tells us how the function's value changes as we move along the x-axis. It's like a roadmap to understanding the slopes of tangent lines to our function at any point. By taking the first derivative, we can identify the rates of change at specific points. In calculus, this is pivotal in finding critical points, where the function might reach a peak or a valley.

To find the first derivative for the function \(y = x^3 - 2x + 4\), we differentiate each term. This step involves using the power rule of differentiation for each term separately:

• The derivative of \(x^3\) is \(3x^2\).
• The derivative of \(-2x\) is \(-2\).
• The derivative of a constant, like 4, is zero.

Thus, the first derivative is \(y' = 3x^2 - 2\). By finding where this derivative equals zero, we unveil critical points which may be potential locations for extreme values of the function.

Exploring the Second Derivative

After we've identified the potential critical points using the first derivative, the second derivative is our tool for digging deeper. It helps classify these critical points as relative maxima, minima, or points of inflection. The key is in the sign of the second derivative at these critical points.

For the function \(y = x^3 - 2x + 4\), let's take another derivative of our first derivative \(3x^2 - 2\):

• The derivative of \(3x^2\) is \(6x\).

Therefore, the second derivative is \(y'' = 6x\). By substituting the critical points from the first derivative solution back into this second derivative equation, we determine the nature of these points:

• If \(y'' > 0\) at a point, the function has a local minimum there.
• If \(y'' < 0\), it finds a local maximum.
• If \(y'' = 0\), further analysis is required as it might be a saddle point.

This analysis aids in understanding the shape of the function curve at critical junctures.

Decoding Critical Points

Critical points are key to understanding where a function's graph changes direction, either peaking or dipping. These points occur where the first derivative is zero or undefined, which implies the tangent line to the graph is horizontal.

For our function \(y = x^3 - 2x + 4\), when we set the first derivative \(3x^2 - 2\) to zero, we solve the equation \(3x^2 - 2 = 0\):

• By solving, we find \(x = \pm \sqrt{\frac{2}{3}}\).

These values \(x = \sqrt{\frac{2}{3}}\) and \(x = -\sqrt{\frac{2}{3}}\) are our critical points.

Utilizing the second derivative \(6x\), we classify these points:

• At \(x = \sqrt{\frac{2}{3}}\), \(y'' > 0\), indicating a local minimum.
• At \(x = -\sqrt{\frac{2}{3}}\), \(y'' < 0\), indicating a local maximum.

These findings help predict where the function curve might be flat or change slope direction.