One of the difficult types of math problems in university entrance exams is finding the maximum (MAX) and minimum (MIN) values of functions. However, for single-variable functions, finding MAX and MIN is quite simple, we just need to master some basic methods to be able to find them. In this article, we will learn what MAX, MIN are and how to find MAX, MIN of common single-variable functions.
See also: $Casio tips$ Finding maximum and minimum values of functions
Definitions of maximum and minimum values
Let y = f(x) be a function defined on the set D.
- The number M is called the maximum value of the function f(x) on D if $f\left( x \right) \le M\,\,\forall x \in D$ and $\exists {x_0} \in D$ such that $f\left( {{x_0}} \right) = M$, denoted as: $\mathop {\max }\limits_D y = M$
- The number m is called the minimum value of the function f(x) on D if $f\left( x \right) \ge m\,\,\forall x \in D$ and $\exists {x_0} \in D$ such that $f\left( {{x_0}} \right) = m$, denoted as: $\mathop {\min }\limits_D y = m$
We can understand that: the largest number among all $f\left( x \right)$ values with $x \in D$ is called MAX and the smallest number among all $f\left( x \right)$ values with $x \in D$ is called MIN.
Methods for finding maximum and minimum values of functions
Method 1: using function variation table. This is a general method for problems of finding maximum and minimum values of functions. Follow these steps:
- Find the domain of the function.
- Find y’, solve y’ = 0.
- Construct a variation table, conclude based on the table.
Example 1: Find the maximum and minimum values of the function $y = x – \sqrt {x – 4} $
Solution
Domain: $D = \left[ {4; + \infty } \right)$
$y’ = 1 – \dfrac{1}{{2\sqrt {x – 4} }}$
$y’ = 0 \Leftrightarrow 1 – \dfrac{1}{{2\sqrt {x – 4} }} = 0 \Leftrightarrow \sqrt {x – 4} = \dfrac{1}{2} \Leftrightarrow x – 4 = \dfrac{1}{4} \Leftrightarrow x = \dfrac{{17}}{4}$
Variation table:
Observation: from the table we see that the minimum value of the function is $\dfrac{{15}}{4}$ and there is no maximum value because the function increases to ${ + \infty }$.
So $\mathop {\min }\limits_{\left[ {4; + \infty } \right)} y = \dfrac{{15}}{4}$ at $x = \dfrac{{17}}{4}$
The function has no maximum value.
See also: Monotonicity of functions and common problem types
Method 2: applied to find maximum and minimum values of function y = f(x) on $a, b$. Follow these steps:
- Find the domain of the function.
- Find y’.
- Find points ${x_1},{x_2},…{x_n}$ in (a,b) where y’ = 0 or y’ is undefined.
- Calculate $f\left( a \right),f\left( b \right),f\left( {{x_1}} \right),f\left( {{x_2}} \right)…f\left( {{x_n}} \right)$
- Conclusion: $\mathop {\max }\limits_{\left[ {a,b} \right]} f\left( x \right) = \max \left\{ {f\left( a \right),f\left( b \right),f\left( {{x_1}} \right),f\left( {{x_2}} \right)…f\left( {{x_n}} \right)} \right\}$ and $\mathop {{\mathop{\rm mim}\nolimits} }\limits_{\left[ {a,b} \right]} f\left( x \right) = \min \left\{ {f\left( a \right),f\left( b \right),f\left( {{x_1}} \right),f\left( {{x_2}} \right)…f\left( {{x_n}} \right)} \right\}$.
Example 2: Find the maximum and minimum values of the function $f\left( x \right) = x + \dfrac{4}{x}$ on $1,3$.
Solution
Domain: $D = \left[ {4; + \infty } \right)$
$f'(x) = 1 – \dfrac{4}{{{x^2}}} $
$f'(x) = 0 \Leftrightarrow 1 – \dfrac{4}{{{x^2}}} = 0 \Leftrightarrow \left[ \begin{array}{l}x = 2 \in \left( {1;3} \right)\\x = – 2 \notin \left( {1;3} \right)\end{array} \right.$
$f\left( 1 \right) = 5,f\left( 2 \right) = 4,f\left( 3 \right) = \dfrac{{13}}{3}$
So $\mathop {\max }\limits_{\left[ {1;3} \right]} f\left( x \right) = 5$ at x = 1, $\mathop {\min }\limits_{\left[ {1;3} \right]} f\left( x \right) = 4$ at x = 2.
Note: some problems only require finding maximum and minimum values of a function without specifying the interval, but if the domain of that function is an interval, we can still use method 2.
Example 3: Find the maximum and minimum values of the function $y = \dfrac{1}{2}\left( {x + \sqrt {4 – {x^2}} } \right)$
Solution
Domain: $D = \left[ { – 2;2} \right]$
$y’ = \dfrac{{\sqrt {4 – {x^2}} – x}}{{2\sqrt {4 – {x^2}} }}$
$y’ = 0 \Leftrightarrow \sqrt {4 – {x^2}} = x \Leftrightarrow \left\{ \begin{array}{l}x \ge 0\\4 – {x^2} = {x^2}\end{array} \right. \Leftrightarrow x = \sqrt 2 \in \left( { – 2;2} \right)$
$y\left( { – 2} \right) = – 1; y\left( {\sqrt 2 } \right) = \sqrt 2 ; y\left( 2 \right) = 1.$
So $\mathop {\max }\limits_{\left[ { – 2;2} \right]} y = y\left( {\sqrt 2 } \right) = \sqrt 2 ;\mathop {\min }\limits_{\left[ { – 2;2} \right]} y = y\left( { – 2} \right) = – 1.$
The above are two basic methods for finding maximum and minimum values of functions that students must master well. This is the foundation to be able to solve more complex problems.
See also: Tangent problems of function graphs
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