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 \( f \).
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 \( m \le f(x) \le \ M \) 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 SolutionsHow to find absolute minimum and absolute maximum of a function \( f \):
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
\( x(x1)(x+2) =0 \)
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
\( f(x) = \dfrac{(x^22x3)(2x2)}{\sqrt{(x^22x3)^2}}1 = \dfrac{(x^22x3)(2x2)  x^22x3}{x^22x3} \)
Step  2: Find the critical points
The denominator of \( f '(x) \) is equal to zero if
\( x^22 x  3 = 0 \)
Factor and solve
\( x^22 x  3 = (x+1)(x3) = 0\)
\( x = 1 \) 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: \( (x^22x3)(2x2)  x^22x3 = 0 \) equation (1)
Expression \( x^22x3 \) depends on \( x \) and may therefore be negative or positive.
If \( x^22x3 < 0 \), \( x^22x3 =  (x^22x3) \) and equation (1) may be written as
\( (x^22x3)(2x2) + (x^22x3) = 0 \)
\( (x^22x3)(2x2+1) = (x^22x3)(2x1) = 0 \)
The above equation has three zeros: \( x = 1\), \( x = 3 \) and \( x = 1/2 \).
If \( x^22x3 > 0 \), \( x^22x3 = (x^22x3) \) and equation (1) may be written as
\( (x^22x3)(2x2)  (x^22x3) = (x^22x3)(2x21) = (x^22x3)(2x3) = 0 \)
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: \( (x^22x3)(2x2)  x^22x3 = ((1/2)^22(1/2)3)(2(1/2)2)  (1/2)^22(1/2)3 = 0 \)
Check \( x = 3/2 \): Numerator: \( (x^22x3)(2x2)  x^22x3 = ((3/2)^22(3/2)3)(2(3/2)2)  (3/2)^22(3/2)3 = 15/2 \)
\( x = 1/2 \) 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 \( f(x) = (x2)^{2/5} \;\; \text{on} \;\; [3 , 4] \).
Solution to Example 8
Step  1: Find the first derivative of \( f \)
\( f '(x) = (2/5)(x2)^{2/51} = (2/5)(x2)^{3/5} = \dfrac{2}{5(x2)^{3/5}} \)
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 LinksCritical Numbers of a Function
