How to find absolute maximum and minimum on a graph

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.

Review of Absolute Minimum and Maximum of a 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 \( f \).

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 \( m \le f(x) \le \ M \) 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.

Examples with Detailed Solutions

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 \( f(x) = - x^2 + 2 x -2 \;\; \text{on} \;\; [-2 , 3] \).

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 \( f(x) = \dfrac{1}{4} x^4 + \dfrac{1}{3} x^3 - x^2 \;\; \text{on} \;\; [-1 , 1] \).

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 \( f(x) = x^2 \ln(x) - 1 \;\; \text{on} \;\; [0.5 , 2] \).

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 \( f(x) = |x^2 - 2x - 3| - x\;\; \text{on} \;\; [-1.1 , 4] \).

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.

\( x^2-2x-3 < 0 \), \( |x^2-2x-3| = - (x^2-2x-3) \)   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 \( x = 1/2 \).

\( x^2-2x-3 > 0 \), \( |x^2-2x-3| = (x^2-2x-3) \)   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 \( x = 3/2 \).

\( f'(x) = 0\)
\( x = -1\) and \( x = 3 \) are not zero of \( f'(x) \) since \( f'(x) \) is undefined at these values.

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

\( x = 3/2 \): 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.

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 \( f(x) = (x-2)^{2/5} \;\; \text{on} \;\; [-3 , 4] \).

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.

More References and Links

Critical Numbers of a Function