Problem 53 - University Calculus: Early Transcendentals

Question

Find the extreme values (absolute and local) of the function over its natural domain, and where they occur. $$y=2 x^{2}-8 x+9$$

Step-by-Step Solution

Answer

The absolute and local minimum occurs at \( x = 2 \) with \( y = 1 \). No absolute maximum.

Step 1: Find the derivative

To find the extreme values, we first need to find the critical points of the function by taking its derivative. The derivative of the function \( y = 2x^2 - 8x + 9 \) with respect to \( x \) is found using the power rule. The derivative is \( y' = 4x - 8 \).

Step 2: Equate derivative to zero

Now, we need to find the values of \( x \) where the derivative is zero, as these points could represent local maxima or minima. We set the derivative \( 4x - 8 = 0 \) and solve for \( x \).

Step 3: Solve for critical points

Solving the equation \( 4x - 8 = 0 \), we get \( 4x = 8 \). Dividing both sides by 4, we find \( x = 2 \). This is our critical point.

Step 4: Evaluate the second derivative

To determine whether the critical point is a local minimum or maximum, we find the second derivative \( y'' \) of the function. Differentiating \( y' = 4x - 8 \), we get \( y'' = 4 \).

Step 5: Analyze the second derivative

Since the second derivative \( y'' = 4 \) is positive, the critical point \( x = 2 \) is a point of local minimum. For a quadratic function, if the second derivative is positive, the function is concave up, indicating a local minimum.

Step 6: Evaluate the function at the critical point

To find the value of the function at the critical point, substitute \( x = 2 \) into the original function: \( y = 2(2)^2 - 8(2) + 9 \). This simplifies to \( y = 8 - 16 + 9 = 1 \). So, the local minimum is \( y = 1 \) when \( x = 2 \).

Step 7: Determine the nature of extreme values

For quadratics, the local minimum in such a case where the parabola opens upwards is also the absolute minimum because it is the only turning point on the natural domain of the function. Thus, the absolute minimum value of \( y \) occurs at \( x = 2 \) and is \( y = 1 \). Since there is no turning point yielding a local maximum, there isn't an absolute maximum on the domain.

Key Concepts

Critical Points

In calculus, critical points are key in finding where a function reaches its extreme values, which could be either a minimum or maximum. To find these points, we first compute the derivative of the function. The critical points occur where this derivative is zero, or where it's undefined. For our function, the derivative is found to be a straightforward linear expression, specifically,

• Given the function: \[ y = 2x^2 - 8x + 9 \]
• Its derivative is: \[ y' = 4x - 8 \]

Finding the critical point involves setting \( y' = 4x - 8 \) equal to zero. This gives us a clear equation to solve. When solved, we find:

• The critical point is at \( x = 2 \).

This critical point is essential for identifying any extreme values the function might have.

Derivative

The derivative is a fundamental tool in calculus, representing the rate of change of the function with respect to its variable. It helps determine the slope of the function at any given point, and finding where the derivative equals zero can reveal critical points.
In our case, the derivative of the function

• \( y = 2x^2 - 8x + 9 \)

is calculated using the power rule for differentiation. This rule states that

• if \( y = ax^n \), then \( y' = nax^{n-1} \).

Applying this to each term gives us the derivative:

• For \( 2x^2 \), the derivative is \( 2*2x^{2-1} = 4x \).
• For \( -8x \), the derivative is \( -8*1x^{1-1} = -8 \).
• Constant terms like \(+9\) become zero.
The complete derivative is therefore \( y' = 4x - 8 \).
This expression is what we use to find critical points for further analysis.

Concavity

Concavity of a function tells us how it curves, which can indicate the nature of any critical points found in the previous steps. We use the second derivative to assess concavity. If the second derivative is positive, the function is concave up, resembling a smile \(:)\), which indicates a local minimum. Conversely, if it is negative, the function is concave down, resembling a frown \(:(\), indicating a local maximum.
For our function, after differentiating the first derivative,

• The second derivative is \( y'' = 4 \).

Here, the second derivative is constant and positive,

• signifying that the function is indeed concave up everywhere.

As a result, the critical point at \( x = 2 \) is confirmed to be a local minimum, which is also the absolute minimum for this function as it opens upwards without any other turning points.