Finding Global Extrema: Videos & Practice Problems

To find local extrema of a function, identify critical points where the derivative equals zero or does not exist. For example, for the function f(x) = x³ - 12x + 5, the derivative f'(x) = 3x² - 12 gives critical points at x = ±2. Not all critical points are local extrema. To find global extrema, use the Extreme Value Theorem, ensuring the function is continuous on a closed interval. Evaluate the function at critical points and endpoints to determine the global maximum and minimum values.

Critical Points

Understanding local extrema is crucial in calculus, particularly when analyzing functions without their graphs. Local extrema refer to the highest or lowest points in a specific interval of a function. To identify these points, we first need to determine the function's critical points, which are essential for locating local maxima and minima.

Critical points occur where the derivative of a function is either zero or undefined. The derivative represents the slope of the tangent line to the function at any given point. For local extrema, the tangent line is either horizontal (indicating a slope of zero) or nonexistent. This means that at local maxima and minima, the derivative will equal zero or not exist.

To find critical points, we start by calculating the derivative of the function. For example, consider the function \( f(x) = x^3 - 12x + 5 \). Using the power rule, the derivative is calculated as:

\( f'(x) = 3x^2 - 12 \)

Next, we set the derivative equal to zero to find where the slope is horizontal:

\( 3x^2 - 12 = 0 \)

Solving this equation involves adding 12 to both sides:

\( 3x^2 = 12 \)

Then, dividing both sides by 3 gives:

\( x^2 = 4 \)

Taking the square root of both sides results in:

\( x = \pm 2 \)

Thus, we have two critical points at \( x = 2 \) and \( x = -2 \). However, it is important to note that not all critical points correspond to local extrema. For instance, a critical point may have a horizontal tangent but not be a local maximum or minimum. This distinction is vital as it emphasizes that identifying critical points is merely the first step in the process of finding local extrema.

In summary, critical points are foundational in analyzing functions, as they help in determining local and global extrema, among other characteristics. As we progress, we will explore further applications of critical points and practice identifying them in various functions.

Critical Points Example 1

To find the critical points of the function \( f(x) = \frac{x^2}{x - 4} \), we first need to determine where the derivative \( f'(x) \) is equal to zero or does not exist. Since this is a rational function, we will apply the quotient rule for differentiation.

The quotient rule states that if you have a function in the form \( \frac{g(x)}{h(x)} \), the derivative is given by:

\[f'(x) = \frac{h(x)g'(x) - g(x)h'(x)}{(h(x))^2}\]

In our case, \( g(x) = x^2 \) and \( h(x) = x - 4 \). The derivatives are \( g'(x) = 2x \) and \( h'(x) = 1 \). Applying the quotient rule, we have:

\[f'(x) = \frac{(x - 4)(2x) - (x^2)(1)}{(x - 4)^2}\]

Now, simplifying the numerator:

\[f'(x) = \frac{2x^2 - 8x - x^2}{(x - 4)^2} = \frac{x^2 - 8x}{(x - 4)^2}\]

Next, we can factor the numerator:

\[f'(x) = \frac{x(x - 8)}{(x - 4)^2}\]

To find the critical points, we set the numerator equal to zero:

\[x(x - 8) = 0\]

This gives us two solutions:

\[x = 0 \quad \text{and} \quad x = 8\]

Next, we check where the derivative does not exist. The derivative will not exist where the denominator is zero, which occurs when:

\[(x - 4)^2 = 0 \implies x = 4\]

However, we must also consider the original function \( f(x) \). The function is undefined at \( x = 4 \) because it would make the denominator zero. Therefore, \( x = 4 \) is not a critical point.

In conclusion, the critical points of the function \( f(x) = \frac{x^2}{x - 4} \) are:

\[x = 0 \quad \text{and} \quad x = 8\]

Find the critical points of the given function.
$g(x)=\frac13x^3-\frac12x^2-12x$

• A. $x=0,x=12$
• B. $x=-4,x=3$
• C. $x=-6,x=2$
• D. $x=4,x=-3$

Answer: (D)

Find the critical points of the given function.
$f(t)=\frac{6t}{t^2+1}$

• A. $t=0,t=6$
• B. $t=1$
• C. $t=-1,t=1$
• D. No critical points

Answer: (C)

Find the critical points of the given function.
$f(x)=\sqrt{2-x^2}$

• A. $x=0$
• B. $x=0,x=\sqrt{2},x=-\sqrt{2}$
• C. $x=0,x=\sqrt2$
• D. No critical points

Answer: (B)

Critical Points Example 2

To find the critical points of the function \( f(\theta) = 2 \sin(\theta) + \cos^2(\theta) \) over the interval \( [0, 2\pi] \), we first need to compute the derivative \( f'(\theta) \). The derivative of the first term, \( 2 \sin(\theta) \), is \( 2 \cos(\theta) \). For the second term, \( \cos^2(\theta) \), we apply the chain rule, resulting in \( 2 \cos(\theta)(-\sin(\theta)) \). Thus, the derivative can be simplified to:

\( f'(\theta) = 2 \cos(\theta) - 2 \cos(\theta) \sin(\theta) = 2 \cos(\theta)(1 - \sin(\theta)) \).

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

\( 2 \cos(\theta)(1 - \sin(\theta)) = 0 \).

This equation gives us two factors to consider:

1. \( 2 \cos(\theta) = 0 \) leads to \( \cos(\theta) = 0 \). The solutions for this within the interval \( [0, 2\pi] \) are \( \theta = \frac{\pi}{2} \) and \( \theta = \frac{3\pi}{2} \).

2. \( 1 - \sin(\theta) = 0 \) simplifies to \( \sin(\theta) = 1 \). The solution for this is \( \theta = \frac{\pi}{2} \), which overlaps with the previous critical point.

Thus, the critical points of the function \( f(\theta) \) in the specified domain are \( \theta = \frac{\pi}{2} \) and \( \theta = \frac{3\pi}{2} \). Since there are no values of \( \theta \) in the interval where the derivative does not exist, these are the only critical points to consider.

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 \([a, b]\), 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 it has global extrema, we first check its continuity. Since this is a polynomial function, it is continuous everywhere. Next, we confirm that it is defined over a closed interval, such as \([-2, 4]\), which is indeed closed. Therefore, we can conclude that \( f(x) \) has a global maximum and minimum.

Once we establish that global extrema exist, we can find them by following a systematic process. The first step is to identify the critical points of the function. This involves calculating the derivative \( f'(x) \) and setting it equal to zero. For our function, the derivative is:

\( f'(x) = 6x \)

Setting this equal to zero gives us the critical point:

\( 6x = 0 \implies x = 0 \)

Next, we evaluate the function at the critical points and the endpoints of the interval. In our case, we will evaluate \( f(x) \) at \( x = -2 \), \( x = 0 \), and \( x = 4 \):

1. \( f(0) = 3(0)^2 + 1 = 1 \)

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

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

Now, we compare these values to find the global maximum and minimum. The values we calculated are:

• At \( x = 0 \), \( f(0) = 1 \)
• At \( x = -2 \), \( f(-2) = 13 \)
• At \( x = 4 \), \( f(4) = 49 \)

The largest value is \( 49 \), which indicates that the global maximum occurs at \( x = 4 \). Conversely, the smallest value is \( 1 \), indicating that the global minimum occurs 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 finding the derivative, and evaluate the function at both the critical points and the endpoints of the interval. This methodical approach ensures that you accurately determine the global maximum and minimum values of the function.

Finding Global Extrema (Extreme Value Theorem) Example 3

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 an absolute minimum. Since \( f(x) \) is a polynomial function, 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 differentiate:

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

Factoring out the common terms gives us:

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

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

\( 8x = 0 \) leads to \( x = 0 \),

\( x - 4 = 0 \) leads to \( x = 4 \),

\( x + 1 = 0 \) leads to \( x = -1 \).

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:

For \( f(0) \):

\( f(0) = 2(0)^4 - 8(0)^3 - 16(0)^2 + 3 = 3 \)

For \( f(4) \):

\( f(4) = 2(4)^4 - 8(4)^3 - 16(4)^2 + 3 = -253 \)

For \( f(-1) \):

\( f(-1) = 2(-1)^4 - 8(-1)^3 - 16(-1)^2 + 3 = -3 \)

For \( f(-2) \):

\( f(-2) = 2(-2)^4 - 8(-2)^3 - 16(-2)^2 + 3 = 35 \)

For \( f(5) \):

\( f(5) = 2(5)^4 - 8(5)^3 - 16(5)^2 + 3 = -147 \)

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 3

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.

To find the global minimum, we need to identify the critical points of the function. This involves calculating the derivative of the function:

\( y' = 2x + 6 \)

Setting the derivative equal to zero to find critical points:

\( 2x + 6 = 0 \)

\( 2x = -6 \)

\( x = -3 \)

Next, we evaluate the function at this critical point to find the corresponding \( y \)-value:

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

\( = 9 - 18 - 3 \)

\( = -12 \)

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 4

To determine the absolute maximum and minimum values of the function \( f(x) = x + \cos(x) \) on the closed interval \([0, 2\pi]\), we can apply the Extreme Value Theorem, which states that a continuous function on a closed interval will attain both a maximum and a minimum.

First, we need to find the critical points of the function by calculating its derivative. The derivative of \( f(x) \) is given by:

\[f'(x) = 1 - \sin(x)\]

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

\[1 - \sin(x) = 0 \implies \sin(x) = 1\]

Within the interval \([0, 2\pi]\), the sine function equals 1 at:

\[x = \frac{\pi}{2}\]

Now, we will evaluate the function \( f(x) \) at the critical point and the endpoints of the interval, which are \( x = 0 \) and \( x = 2\pi \).

Calculating \( f \) at these points:

1. For \( x = 0 \):\[f(0) = 0 + \cos(0) = 0 + 1 = 1\]2. For \( x = \frac{\pi}{2} \):\[f\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + \cos\left(\frac{\pi}{2}\right) = \frac{\pi}{2} + 0 = \frac{\pi}{2} \approx 1.57\]3. For \( x = 2\pi \):\[f(2\pi) = 2\pi + \cos(2\pi) = 2\pi + 1 \approx 6.28 + 1 = 7.28\]

Now we compare the function values:

- \( f(0) = 1 \)- \( f\left(\frac{\pi}{2}\right) \approx 1.57 \)- \( f(2\pi) \approx 7.28 \)

From these calculations, we can identify the absolute maximum and minimum values:

- The absolute maximum value is \( f(2\pi) = 2\pi + 1 \) at \( x = 2\pi \).- The absolute minimum value is \( f(0) = 1 \) at \( x = 0 \).

In conclusion, the function \( f(x) = x + \cos(x) \) has an absolute maximum of \( 2\pi + 1 \) at \( x = 2\pi \) and an absolute minimum of \( 1 \) at \( x = 0 \).

Here’s what students ask on this topic:

How do you find the global maximum and minimum of a function without a graph?

To find the global maximum and minimum of a function without a graph, follow these steps: First, use the Extreme Value Theorem to ensure the function is continuous on a closed interval. Then, find the 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 obtained is the global maximum, and the lowest value is the global minimum.

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 has both a global maximum and a global minimum on that interval. To use this theorem, first verify that the function is continuous and defined over a closed interval. Then, find the critical points and evaluate the function at these points and the endpoints. The highest and lowest values obtained are the global maximum and minimum, respectively.

What are critical points and how do you find them?

Critical points of a function are where its derivative is zero or does not exist. To find them, take the derivative of the function and set it equal to zero, then solve for x. Additionally, check where the derivative does not exist within the domain of the function. These x-values are the critical points, which are essential for identifying local and global extrema.

Can a function have critical points that are not local extrema?

Yes, a function can have critical points that are not local extrema. Critical points are where the derivative is zero or undefined, but not all such points correspond to local maxima or minima. For example, a critical point might be a point of inflection where the function changes concavity but does not have a local maximum or minimum.

How do you determine if a function has a global maximum or minimum?

To determine if a function has a global maximum or minimum, use the Extreme Value Theorem. Ensure the function is continuous on a closed interval. If it is, the function will have both a global maximum and minimum within that interval. Then, find the critical points and evaluate the function at these points and the endpoints to identify the global extrema.