Finding Global Extrema: Videos & Practice Problems

To find the global maximum and minimum of a function, first verify continuity and that it is defined over a closed interval using the Extreme Value Theorem. For a function like $f^x=3x^2+1$, check critical points by setting the derivative $f\text{'}=6x$ to zero. Evaluate the function at critical points and endpoints to identify the global extrema, with the largest value as the global maximum and the smallest as the global minimum.

Finding Global Extrema (Extreme Value Theorem)

To identify the global maximum and minimum values of a function, we can utilize the Extreme Value Theorem, which states that if a function \( f \) is continuous on a closed interval \([a, b]\), then it must have both a global maximum and a global minimum within that interval. This means that the function should not have any breaks or holes, and both endpoints \( a \) and \( b \) must be included in the interval.

For example, consider the function \( f(x) = 3x^2 + 1 \). To determine if this function has global extrema, we first check its continuity. Since \( f(x) \) is a polynomial, it is continuous everywhere. Next, we confirm that the interval from \(-2\) to \(4\) is closed, as indicated by the square brackets. Therefore, we conclude that \( f(x) \) does indeed have a global maximum and minimum.

Once we establish that global extrema exist, we can find them by identifying critical points and evaluating the function at these points and the endpoints of the interval. Critical points occur where the derivative \( f'(x) \) is zero or undefined. For our function, we calculate the derivative:

$$ f'(x) = 6x $$

Setting the derivative equal to zero gives us:

$$ 6x = 0 \implies x = 0 $$

Next, we evaluate the function at the critical point and the endpoints of the interval. The endpoints are \( x = -2 \) and \( x = 4 \). We compute:

1. For \( x = 0 \):

$$ f(0) = 3(0)^2 + 1 = 1 $$

2. For \( x = -2 \):

$$ f(-2) = 3(-2)^2 + 1 = 3(4) + 1 = 13 $$

3. For \( x = 4 \):

$$ f(4) = 3(4)^2 + 1 = 3(16) + 1 = 49 $$

Now we compare the values obtained: \( f(0) = 1 \), \( f(-2) = 13 \), and \( f(4) = 49 \). The largest value, \( 49 \), represents the global maximum, occurring at \( x = 4 \). Conversely, the smallest value, \( 1 \), indicates the global minimum, occurring at \( x = 0 \).

In summary, to find the global extrema of a function, verify the conditions of the Extreme Value Theorem, identify critical points by setting the derivative to zero, and evaluate the function at these points and the endpoints of the interval. This systematic approach ensures accurate identification of the global maximum and minimum values.

Finding Global Extrema (Extreme Value Theorem) Example 1

To find the absolute maximum and minimum values of the function \( f(x) = 2x^4 - 8x^3 - 16x^2 + 3 \) over the closed interval \([-2, 5]\), we can apply the Extreme Value Theorem. This theorem states that a continuous function on a closed interval will attain both an absolute maximum and minimum. Since \( f(x) \) is a polynomial, it is continuous everywhere, confirming that it will have both values on the specified interval.

The first step is to find the critical points by calculating the derivative \( f'(x) \) and setting it equal to zero. Using the power rule, we find:

\( f'(x) = 8x^3 - 24x^2 - 32x \)

Factoring out the common terms gives:

\( f'(x) = 8x(x^2 - 3x - 4) = 8x(x - 4)(x + 1) \)

Setting each factor to zero, we find the critical points:

• \( 8x = 0 \) → \( x = 0 \)
• \( x - 4 = 0 \) → \( x = 4 \)
• \( x + 1 = 0 \) → \( x = -1 \)

Thus, the critical points are \( x = 0, 4, -1 \). Next, we evaluate the function at these critical points and the endpoints of the interval, \( x = -2 \) and \( x = 5 \).

Calculating the function values:

• \( f(0) = 2(0)^4 - 8(0)^3 - 16(0)^2 + 3 = 3 \)
• \( f(4) = 2(4)^4 - 8(4)^3 - 16(4)^2 + 3 = -253 \)
• \( f(-1) = 2(-1)^4 - 8(-1)^3 - 16(-1)^2 + 3 = -3 \)
• \( f(-2) = 2(-2)^4 - 8(-2)^3 - 16(-2)^2 + 3 = 35 \)
• \( f(5) = 2(5)^4 - 8(5)^3 - 16(5)^2 + 3 = -147 \)

Now we compile the function values:

• \( f(-2) = 35 \)
• \( f(-1) = -3 \)
• \( f(0) = 3 \)
• \( f(4) = -253 \)
• \( f(5) = -147 \)

From these values, we identify the absolute maximum and minimum. The largest value is \( 35 \) at \( x = -2 \), indicating that the global maximum is \( 35 \). The smallest value is \( -253 \) at \( x = 4 \), indicating that the global minimum is \( -253 \).

In conclusion, the absolute maximum value of the function on the interval \([-2, 5]\) is \( 35 \) at \( x = -2 \), and the absolute minimum value is \( -253 \) at \( x = 4 \).

Find the global maximum and minimum values of the function on the given interval. State as ordered pairs.
$y=x+\frac{2}{x};[0.25,3]$

• A. Global maximum at $\left(0.25,8.25\right)$, Global minimum at $\left(0,2\right)$
• B. Global maximum at $\left(0.25,8.25\right)$, Global minimum at $\left(\sqrt2,2\sqrt2\right)$
• C. Global maximum at $\left(0.25,8.25\right)$, Global minimum at $\left(3,3.67\right)$
• D. Global maximum at $\left(\sqrt2,2\sqrt2\right)$, Global minimum at $\left(-\sqrt2,-2\sqrt2\right)$

Answer: (B)

Find the global maximum and minimum values of the function on the given interval. State as ordered pairs.
$y=8+27x-x^3;[0,4]$

• A. Global maximum at $\left(3,62\right)$, Global minimum at $\left(0,8\right)$
• B. Global maximum at $\left(3,62\right)$, Global minimum at $\left(4,52\right)$
• C. Global maximum at $\left(-3,-46\right)$, Global minimum at $\left(0,8\right)$
• D. Global maximum at $\left(4,52\right)$, Global minimum at $\left(-3,-46\right)$

Answer: (A)

Finding Global Extrema (Extreme Value Theorem) Example 2

To determine the global maximum and minimum values of the function \( y = x^2 + 6x - 3 \) over the interval from negative infinity to positive infinity, we first recognize that this is a quadratic function represented by a parabola. Since the coefficient of \( x^2 \) is positive, the parabola opens upwards, indicating that it will not have a global maximum, as both ends extend to infinity.

However, the function will have a global minimum, which occurs at the vertex of the parabola. To find this minimum, we need to locate the critical points by taking the derivative of the function. The derivative is calculated as follows:

\( y' = 2x + 6 \)

Setting the derivative equal to zero to find critical points:

\( 2x + 6 = 0 \)

\( 2x = -6 \)

\( x = -3 \)

Now that we have the critical point \( x = -3 \), we substitute this value back into the original function to find the corresponding \( y \)-value:

\( y = (-3)^2 + 6(-3) - 3 \)

\( y = 9 - 18 - 3 \)

\( y = -12 \)

Thus, the global minimum value of the function is \( -12 \) at \( x = -3 \). Since the parabola continues to rise indefinitely, there is no global maximum value. Therefore, we conclude that the function has a global minimum of \( -12 \) at \( x = -3 \) and no global maximum.

Finding Global Extrema (Extreme Value Theorem) Example 3

To find the absolute maximum and minimum values of the function \( f(x) = x \ln(x) \) over the closed interval \([1, 2]\), we can apply the Extreme Value Theorem. This theorem states that a continuous function on a closed interval will attain both a maximum and a minimum value.

First, we need to identify the critical points of the function by calculating its derivative. Using the product rule, which states that the derivative of a product \( u \cdot v \) is given by \( u'v + uv' \), we differentiate \( f(x) \). Here, let \( u = x \) and \( v = \ln(x) \). Thus, we have:

\( f'(x) = u'v + uv' = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1 \)

Next, we set the derivative equal to zero to find critical points:

\( \ln(x) + 1 = 0 \implies \ln(x) = -1 \implies x = e^{-1} = \frac{1}{e} \approx 0.37 \)

However, since \( \frac{1}{e} \) is not within the interval \([1, 2]\), we will only evaluate the function at the endpoints of the interval.

Now, we calculate the function values at the endpoints:

1. For \( x = 1 \):

\( f(1) = 1 \cdot \ln(1) = 1 \cdot 0 = 0 \)

2. For \( x = 2 \):

\( f(2) = 2 \cdot \ln(2) \approx 2 \cdot 0.693 = 1.386 \)

Now we compare the values obtained:

• At \( x = 1 \), \( f(1) = 0 \) (minimum value)
• At \( x = 2 \), \( f(2) \approx 1.386 \) (maximum value)

Thus, the absolute minimum value of the function on the interval \([1, 2]\) is \( 0 \) at \( x = 1 \), and the absolute maximum value is approximately \( 1.386 \) at \( x = 2 \).

Here’s what students ask on this topic:

What is the Extreme Value Theorem and how is it used to find global extrema?

The Extreme Value Theorem states that if a function is continuous on a closed interval [a, b], then it must have both a global maximum and a global minimum on that interval. To use this theorem for finding global extrema, first ensure the function is continuous and defined over a closed interval. Then, identify critical points by setting the derivative equal to zero and solving for x. Evaluate the function at these critical points and at the endpoints of the interval. The highest value found is the global maximum, and the lowest value is the global minimum. This process ensures that you identify the absolute highest and lowest points of the function within the specified interval.

How do you find critical points when determining global extrema?

To find critical points when determining global extrema, you need to take the derivative of the function and set it equal to zero. This will help you find the x-values where the slope of the tangent is zero, indicating potential maxima or minima. For example, if you have a function f(x) = 3x² + 1, the derivative is f'(x) = 6x. Setting this equal to zero gives x = 0 as a critical point. Additionally, check where the derivative does not exist, although for polynomials like this one, the derivative exists everywhere. Once you have the critical points, evaluate the function at these points and at the endpoints of the interval to find the global extrema.

Why is it important to evaluate a function at endpoints when finding global extrema?

Evaluating a function at endpoints is crucial when finding global extrema because the highest or lowest values of a function on a closed interval can occur at these points. The Extreme Value Theorem guarantees that a continuous function on a closed interval will have a global maximum and minimum, but these extrema can be located at critical points or at the endpoints. By evaluating the function at both critical points and endpoints, you ensure that you identify the absolute highest and lowest values within the interval. This comprehensive approach prevents missing any potential extrema that might occur at the boundaries of the interval.

Can a function have a global maximum or minimum if it is not continuous?

A function that is not continuous may not have a global maximum or minimum, especially if it has breaks, holes, or asymptotes within the interval. The Extreme Value Theorem applies only to continuous functions on closed intervals, ensuring the existence of global extrema. If a function is discontinuous, it might not reach a highest or lowest point, as seen in functions with vertical asymptotes or jumps. However, in some cases, a discontinuous function might still have global extrema if the discontinuities do not affect the overall highest or lowest values within the interval. Careful analysis is required to determine the presence of global extrema in such cases.

What role does the derivative play in finding global extrema?

The derivative plays a crucial role in finding global extrema by helping identify critical points where the function's slope is zero or undefined. These points are potential locations for local maxima or minima. By taking the derivative of a function and setting it equal to zero, you can solve for x-values where the function's rate of change is zero, indicating a possible peak or trough. Additionally, the derivative helps determine where the function is increasing or decreasing, providing insight into the behavior of the function around critical points. Evaluating the function at these critical points and endpoints allows you to find the global maximum and minimum values within a specified interval.