Maxima and Minima

Maxima and minima are used to find the highest or lowest value of a quantity, such as greatest profit, shortest distance, maximum area, or minimum cost.

This topic forms an important part of Applications of Derivatives and connects graph interpretation, increasing-decreasing behavior, and optimization problems.

Let f be a function defined on an interval I.

Maximum value: f has a maximum value at c ∈ I if \[ \boxed{f(c) \geq f(x) \quad \text{for all } x \in I} \]

The value f(c) is called the maximum value and c is called a point of maximum.

Minimum value: f has a minimum value at c ∈ I if \[ \boxed{f(c) \leq f(x) \quad \text{for all } x \in I} \]

The value f(c) is called the minimum value and c is called a point of minimum.

Extreme value: A maximum or minimum value of f is called an extreme value.

 
           Maximum value                          Minimum value

A point in the domain of a function is called a critical point if either the derivative is zero there or the derivative does not exist there. Critical points are checked while locating possible maxima or minima.

The points where a function changes from decreasing to increasing or from increasing to decreasing are called turning points.

      Local Maximum at x = c                            Local Minimum at x = c  

Let c be a critical point of a continuous function f:

• Local Maximum: If f'(x) changes sign from positive to negative as x passes through c, then cc is a point of local maximum.

• Local Minimum: If f'(x) changes sign from negative to positive then c is a point of local minimum.

• Point of Inflection: f'(x) does not change sign as x passes through c (it is neither a maxima nor a minima).

Assume f'(c) = 0 and the second derivative exists at c:

• Local Maximum: f''(c) < 0

• Local Minimum: f''(c) > 0

• Test Fails: f''(c) = 0. If this happens, you must go back and use the First Derivative Test to check if it is a maxima, minima, or point of inflection.

For a continuous function on a closed interval \([a,b]\), absolute maxima and minima are found by comparing function values at:

• critical points inside the interval,

• the left endpoint \(a\), and

• the right endpoint \(b\).

This comparison is necessary because the largest or smallest value may occur at an endpoint.

Find local maximum and local minimum values of the function f given by f (x) = 3x⁴ + 4x³ – 12x² + 12

Solution: 

First find the derivative: \[ f'(x) = 12x^3 + 12x^2 - 24x \]

\[ = 12x(x - 1)(x + 2) \]

For critical points, \[ f'(x) = 0 \]

So, \[ x = 0, \quad x = 1, \quad x = -2 \]

Now find the second derivative: \[ f''(x) = 36x^2 + 24x - 24 \]

At x = 0:

\[ f''(0) = -24 < 0 \]

Therefore, x = 0 is a point of local maximum.

\[ f(0) = 12 \]

Hence, \[ \boxed{\text{Local maximum value} = 12} \]

At x = 1:

\[ f''(1) = 36 > 0 \]

Therefore, x = 1 is a point of **local minimum**.

\[ f(1) = 3 + 4 - 12 + 12 = 7 \]

Hence, \[ \boxed{\text{Local minimum value} = 7} \]

At x = -2:

\[ f''(-2) = 72 > 0 \]

Therefore, x = -2 is also a point of **local minimum**.

\[ f(-2) = 48 - 32 - 48 + 12 = -20 \]

Hence, \[ \boxed{\text{Local minimum value} = -20} \]

Final Answer:

\[ \boxed{\text{Local maximum value} = 12 \text{ at } x = 0} \]

\[ \boxed{\text{Local minimum values} = 7 \text{ at } x = 1 \text{ and } -20 \text{ at } x = -2} \]

• Maxima and minima are extreme values of a function.

• Critical points occur where \(f'(x)=0\) or \(f'(x)\) is not defined.

• If \(f'(x)\) changes from positive to negative, the function has a local maximum.

• If \(f'(x)\) changes from negative to positive, the function has a local minimum.

• If \(f''(c) < 0\), there is a local maximum at \(x=c\).

• If \(f''(c) > 0\), there is a local minimum at \(x=c\).

• For absolute extrema on \([a,b]\), compare values at critical points and endpoints.

• Not every critical point gives a maximum or minimum.

• The second derivative test is quick, but the first derivative test is often more reliable in detailed reasoning.