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# Maxima and Minima

Maxima and minima are the highest and lowest values, respectively, that a function or a set of data can reach. In calculus, maxima and minima are important concepts in optimization problems and in finding critical points of functions.

• Maxima and Minima
• Formula for Maxima
• Solved Examples for Maxima and Minima
• FAQs

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## Maxima and Minima

Here are some key points to keep in mind about maxima and minima:

A function can have multiple maxima and minima, or none at all.

The maxima and minima of a function occur at critical points, which are points where the derivative of the function is zero or undefined.

The maxima and minima of a function can also occur at endpoints of a closed interval, such as in a constrained optimization problem.

To find the maxima and minima of a function, you can use techniques such as the first and second derivative tests, or optimization methods such as linear programming or dynamic programming.

In summary, maxima and minima are important concepts in calculus and optimization, and they can help us find optimal solutions to problems in various fields, such as economics, engineering, and physics. ## Formula for Maxima

The formula for finding maxima of a function using calculus involves finding the critical points of the function and then evaluating the second derivative of the function at those points. If the second derivative is negative, then the critical point is a local maximum. The formula can be written as: Find the critical points of the function f(x) by solving f'(x) = 0 or f'(x) does not exist.
Evaluate the second derivative of the function f”(x) at each critical point.
If f”(x) < 0 at a critical point x, then f(x) has a local maximum at x.
Alternatively, the function can also have a global maximum at one of the endpoints of the interval where the function is defined.

Note that finding the maxima of a function can be more complex for multivariable functions, where partial derivatives and Hessian matrices are used to find critical points and determine the nature of those critical points.

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## Solved Examples for Maxima & Minima

Here are some examples of finding maxima and minima of functions:

Example 1: Find the local maxima and minima of the function f(x) = x^3 – 3x^2 + 2x.

Solution: To find the critical points, we need to find where the derivative is zero or undefined:
f'(x) = 3x^2 – 6x + 2 = 0
Solving for x, we get x = 1 ± sqrt(2)/3.
These are the critical points. Now we need to evaluate the second derivative at each point to determine whether it is a local maximum or minimum:
f”(x) = 6x – 6
At x = 1 – sqrt(2)/3, f”(x) < 0, so this is a local maximum.
At x = 1 + sqrt(2)/3, f”(x) > 0, so this is a local minimum.
Therefore, the function has a local maximum at x = 1 – sqrt(2)/3 and a local minimum at x = 1 + sqrt(2)/3.

Example 2: Find the global maximum and minimum of the function f(x) = x^2 – 4x + 3 on the interval [0, 3].

Solution: Since the interval is closed, we also need to check the endpoints.
f(0) = 3
f(3) = 0
To find the critical points, we need to find where the derivative is zero or undefined:
f'(x) = 2x – 4 = 0
Solving for x, we get x = 2.
This is the only critical point in the interval [0, 3]. Now we need to evaluate the second derivative at this point to determine whether it is a global maximum or minimum:
f”(x) = 2 > 0, so x = 2 is a global minimum.
Therefore, the function has a global minimum of -1 at x = 2, and a global maximum of 3 at x = 0.

Example 3: Find the maximum and minimum values of the function f(x) = x^3 – 6x^2 + 9x + 2 on the interval [-1, 4].

Solution: We start by checking the endpoints of the interval:
f(-1) = 10
f(4) = 10
Next, we find the critical points by finding where the derivative is zero or undefined:
f'(x) = 3x^2 – 12x + 9 = 0
Solving for x, we get x = 1 or x = 3.
These are the critical points. Now we need to evaluate the second derivative at each point to determine whether it is a maximum or minimum:
f”(x) = 6x – 12
At x = 1, f”(x) < 0, so this is a local maximum.
At x = 3, f”(x) > 0, so this is a local minimum.
Therefore, the function has a local maximum of 10 at x = -1, a local maximum of 13 at x = 1, and a local minimum of -8 at x = 3.
Thus, the maximum value of the function on the interval [-1, 4] is 13, and the minimum value is -8.

## Maxima and Minima FAQS

##### What is the difference between local maxima/minima and global maxima/minima?

Local maxima/minima are points on a graph where the function reaches its highest or lowest point in a local area, while global maxima/minima are the highest or lowest point on the entire graph or an interval. A local maximum is higher than all nearby points, but there may be a higher point further away. Similarly, a local minimum is lower than all nearby points, but there may be a lower point further away. A global maximum or minimum is the highest or lowest point on the entire graph or an interval.

##### How do you find the critical points of a function?

The critical points of a function are the points where the derivative of the function is zero or undefined. To find the critical points, you need to take the derivative of the function, set it equal to zero, and solve for x. If the derivative is undefined at a point, you should also check whether the function has a vertical tangent at that point.

##### How do you determine whether a critical point is a maximum or minimum?

To determine whether a critical point is a maximum or minimum, you need to evaluate the second derivative of the function at that point. If the second derivative is positive, the critical point is a local minimum. If the second derivative is negative, the critical point is a local maximum. If the second derivative is zero, the test fails and you need to use another method to determine whether the point is a maximum or minimum.

##### What is a stationary point?

A stationary point is a point on a graph where the slope of the tangent is zero. Stationary points include local maxima/minima and points where the function changes direction (inflection points). Gloria Mathew writes on math topics for K-12. A trained writer and communicator, she makes math accessible and understandable to students at all levels. Her ability to explain complex math concepts with easy to understand examples helps students master math. LinkedIn

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