# Absolute Minimum and Maximum of a Function

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 $f$.

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 Solutions

How 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
$x(x-1)(x+2) =0$
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.