The absolute minimum and maximum of a function are studied using graphs and examples with detailed solutions. Graphical solutions to the examples are also presented for deep understanding of the absolute minimum and maximum of a given function.

__Definition:__ Values of x in the domain of function f at which f '(x) = 0 or f '(x) is undefined are called critical
points of function f .

__Theorem:__ For a continous function on a closed interval [ a,b ], there always exist values x_{1} and x_{2} in [ a,b ] such that f(x_{1}) = m is an absolute minimum and f(x_{2}) = M is an absolute maximum or m ? f(x) ? M for every x in [ a,b ].

Examples on absolute maximum and minimum of functions in different situations are discussed graphically.

Example 1 Absolute minimum and maximum at stationary points

Absolute minimum and maximum of a function may happen at local minimum and maximum respectively as shown in the graph below. Local minima and maxima of a function occur at values of x = x_0 included in the domain of f
such that f '(x_{0}) = 0 and f '(x) changes sign at x = x_{0} . In the system of coordinates below are shown function f (blue) and its derivative (red).
We can see that f has an absolute minimum at x = x_{1} where f '(x) = 0 and changes sign from negative to positive and has an absolute maximum at x = x_{2} where f '(x_{2}) = 0 changes sign from positive to negative.

Example 2 Absolute minimum and maximum at points where the first derivative is undefined

The absolute minimum and maximum of a function may occur at points where the first derivative is undefined as shown in the graph below.

Example 3 Absolute minimum and maximum at end points

The absolute minimum and maximum of a function may happen at the endpoints of the interval defining the domain of the function.

In the example below, \( f(x) = x^3 - 2x^2 \) for \( -1\le x \le 5/2 \) where \( -1 \) and \( 5/2 \) are the endpoints of the interval \( [-1,5/2] \) defining the domain of the function. Note that although there is a local minimum and a local maximum, they are not absolute minimum and maximum.

Step 1: Find the first derivative of function \( f \)

Step 2: Find the critical points of function \( f \)

Step 3: Evaluate \( f \) at all endpoints and critical points and take the smallest (minimum) and largest (maximum) values.

Example 4

Find the absolute maximum and minimum of function \( f \) defined by \( f(x) = - x^2 + 2 x -2 \;\; \text{on} \;\; [-2 , 3] \).

__Solution to Example 4__

Step - 1: Find the first derivative of \( f \)

\( f '(x) = -2 x + 2 \)

Step - 2: Find the critical points (see definition above) of the derivative

Zeros: \( - 2 x + 2 = 0 \)

x = 1 is the zero of \( f '(x) \)

The first derivative is defined everywhere within the domain of the function given by \( [-2 , 3] \) and its zero at \( x = 1 \) is within the domain.

Hence function \( f \) has one critical point at x = 1

Step - 3: Evaluate the function at the endpoints of the interval \( [-2 , 3] \) and the critical points

\( f(-2) = - (-2)^2 + 2 (-2) -2 = -10\)

\( f(3) = - (3)^2 + 2 (3) -2 = -5 \)

\( f(1) = - (1)^2 + 2 (1) - 2 = - 1 \)

The absolute maximum value of \( f(x) \) is: \( -1 \) at \(x = -1\)

The absolute minimum value of \( f(x) \) is: \( -10 \) at \(x = -2\)

The graph of \( f \) is shown below with the critical point and endpoints as well the absolute minimum and absolute maximum.

Example 5

Find the absolute maximum and minimum of function \( f \) defined by \( f(x) = \dfrac{1}{4} x^4 + \dfrac{1}{3} x^3 - x^2 \;\; \text{on} \;\; [-1 , 1] \).

__Solution to Example 5__

Step - 1: Find the first derivative of the given function \( f \)

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

Step - 2: Find the critical points

Zeros: \( x^3 + x^2 - 2x = 0 \)

Factor left hand side

\( x(x-1)(x+2) =0 \)

Zeros of the derivative \( f ' \) are: \( x = 0 \) , \( x = 1 \) and \( x = - 2\)

The first derivative is defined everywhere within the domain of the function given by \( [-1 , 1] \) and only two zero \( x = 0 \) and \( x = 1 \) are within the domain.

Hence function \( f \) has two critical points within the domain: at \( x = 0 \) and \( x = 1 \)

Step - 3: Evaluate the function at the endpoints of the interval \( [-1 , 1] \) and the critical points \( x = 0 \) and \( x = 1 \).

\( f(-1) = \dfrac{1}{4} (-1)^4 + \dfrac{1}{3} (-1)^3 - (-1)^2 = - 13/12\)

\( f(1) = \dfrac{1}{4} (1)^4 + \dfrac{1}{3} (1)^3 - (1)^2 = -5/12\)

\( f(0) = \dfrac{1}{4} (0)^4 + \dfrac{1}{3} (0)^3 - (0)^2 = 0 \)

The absolute maximum value of \( f(x) \) is: \( 0 \) at \( x = 0 \)

The absolute minimum value of \( f(x) \) is: \( -13/12 \) at \( x = - 1\)

The graph of \( f \) is shown below with the critical points and endpoints as well the absolute minimum and absolute maximum.

Example 6

Find the absolute maximum and minimum of function \( f \) defined by \( f(x) = x^2 \ln(x) - 1 \;\; \text{on} \;\; [0.5 , 2] \).

__Solution to Example 4__

Step - 1: Find the first derivative of \( f \)

\( f '(x) = 2x \ln(x) + x^2 (1/x) = 2x \ln(x) + x \)

Step - 2: Find the critical points

Zeros: \(2x \ln(x) + x = 0 \)

Factor left hand side

\( x( 2\ln(x) + 1) =0 \)

Solve the above

\( x = 0 \)

and

\( 2 \ln(x) + 1 = 0 \) which give \( \ln(x) = - 1/2 \) and \( x = e^{-1/2} \)

Zeros of the derivative \( f ' \) are: \( x = 0 \) and \( x = e^{-1/2} \approx 0.61 \)

The first derivative is defined everywhere within the domain of the function given by \( [0.5 , 2] \) and only one zero \( x = e^{-1/2} \) is is within the domain.

Hence function \( f \) has one critical point within the domain: at \( x = e^{-1/2} \)

Step - 3: Evaluate the function at the endpoints of the interval \( [0.5 , 2] \) and the critical point calculated above.

\( f(0.5) = (0.5)^2 \ln(0.5) - 1 = -1.17 \)

\( f(2) = (2)^2 \ln(2) - 1 = 1.77 \)

\( f(e^{-1/2}) = (e^{-1/2})^2 \ln((e^{-1/2})) - 1 = -1.18 \)

The absolute maximum value of \( f(x) \) is: \( 1.77 \) at \( x = 2 \)

The absolute minimum value of \( f(x) \) is: \( -1.18 \) at \( x = e^{-1/2}\)

Example 7

Find the absolute maximum and minimum of function \( f \) defined by \( f(x) = |x^2 - 2x - 3| - x\;\; \text{on} \;\; [-1.1 , 4] \).

__Solution to Example 7__

Step - 1: Find the first derivative of \( f \)

Use the fact that \( \sqrt{u^2} = |u| \) to write \( f(x) \) as follows

\( f(x) = |x^2 - 2x - 3| - x = \sqrt{(x^2 - 2x - 3)^2} - x\)

The first derivative of \( f \) is

\( f(x) = \dfrac{(x^2-2x-3)(2x-2)}{\sqrt{(x^2-2x-3)^2}}-1 = \dfrac{(x^2-2x-3)(2x-2) - |x^2-2x-3|}{|x^2-2x-3|} \)

Step - 2: Find the critical points

The denominator of \( f '(x) \) is equal to zero if

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

Factor and solve

\( x^2-2 x - 3 = (x+1)(x-3) = 0\)

\( x = -1 \) and \( x = 3 \) make the denominator of \(f'(x) \) equal to zero and therefore are critical points because \( f '(x) \) is undefined at these values of \( x \).

Find zeros by setting numerator equal to zero: \( (x^2-2x-3)(2x-2) - |x^2-2x-3| = 0 \) equation (1)

Expression \( x^2-2x-3 \) depends on \( x \) and may therefore be negative or positive.

__If __ \( x^2-2x-3 < 0 \), \( |x^2-2x-3| = - (x^2-2x-3) \) and equation (1) may be written as

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

\( (x^2-2x-3)(2x-2+1) = (x^2-2x-3)(2x-1) = 0 \)

The above equation has three zeros: \( x = -1\), \( x = 3 \) and \( x = 1/2 \).

__If__ \( x^2-2x-3 > 0 \), \( |x^2-2x-3| = (x^2-2x-3) \) and equation (1) may be written as

\( (x^2-2x-3)(2x-2) - (x^2-2x-3) = (x^2-2x-3)(2x-2-1) = (x^2-2x-3)(2x-3) = 0 \)

The above equation has three zeros: \( x = -1\), \( x = 3 \) and \( x = 3/2 \).

__Check solutions of __ \( f'(x) = 0\)

\( x = -1\) and \( x = 3 \) are not zero of \( f'(x) \) since \( f'(x) \) is undefined at these values.

__Check__ \( x = 1/2 \): Numerator: \( (x^2-2x-3)(2x-2) - |x^2-2x-3| = ((1/2)^2-2(1/2)-3)(2(1/2)-2) - |(1/2)^2-2(1/2)-3| = 0 \)

__Check__ \( x = 3/2 \): Numerator: \( (x^2-2x-3)(2x-2) - |x^2-2x-3| = ((3/2)^2-2(3/2)-3)(2(3/2)-2) - |(3/2)^2-2(3/2)-3| = -15/2 \)

\( x = 1/2 \) is the only zero of \( f'(x) \) and hence a critical point.

__Conclusion__ Function \( f \) has three critical points: \( x = 1/2 \) is a zero of \( f'(x) \) and \( x = -1\) and \( x = 3 \) are values of \( x \) for which \( f'(x) \) is undefined.

Step - 3: Evaluate the function at the endpoints of the interval \( [-1.1 , 4] \) and the critical points \( -1 \) , \( 1/2 \) and \( 3 \).

\( f(-1.1) = |(-1.1)^2 - 2(-1.1) - 3| - (-1.1) = 1.51 \)

\( f(4) = |(4)^2 - 2(4) - 3| - (4) = 1 \)

\( f(1/2) = |(1/2)^2 - 2(1/2) - 3| - (1/2) = 13/4 = 3.25 \)

\( f(-1) = |(-1)^2 - 2(-1) - 3| - (-1) = 1 \)

\( f(3) = |(3)^2 - 2(3) - 3| - (3) = -3 \)

The absolute maximum value of \( f(x) \) is: \( 3.25 \) at \( x = 1/2 \)

The absolute minimum value of \( f(x) \) is: \( -3 \) at \( x = 3\)

The graph of \( f \) is shown below with the critical points calculated above and the endpoints of the interval \( [-1.1 , 4] \) as well the absolute minimum and absolute maximum.

Example 8

Find the absolute maximum and minimum of function \( f \) defined by \( f(x) = (x-2)^{2/5} \;\; \text{on} \;\; [-3 , 4] \).

__Solution to Example 8__

Step - 1: Find the first derivative of \( f \)

\( f '(x) = (2/5)(x-2)^{2/5-1} = (2/5)(x-2)^{-3/5} = \dfrac{2}{5(x-2)^{3/5}} \)

Step - 2: Find the critical points (see definition above) of the derivative

\( f'(x) \) does not have any zeros.

The denominator of the first derivative is equal to zero at \( x = 2\) and therefore \( x = 2 \) is a critical point.

Step - 3: Evaluate the function at the endpoints and the critical points

\( f(-3) = ((-3)-2)^{2/5} \approx 1.90\)

\( f(4) = ((4)-2)^{2/5} \approx 1.32\)

\( f(2) = ((2)-2)^{2/5} = 0 \)

The absolute maximum value of \( f(x) \) is: \( \approx 1.90 \) at \(x = -3\)

The absolute minimum value of \( f(x) \) is: \( 0 \) at \(x = 2\)

The graph of \( f \) is shown below with the critical point determined above and the endpoints of the interval \( [-3 , 4] \) as well the absolute minimum and absolute maximum.