Monotonicity, Maxima and Minima
9.0 Absolute Maxima and Absolute Minima
9.0 Absolute Maxima and Absolute Minima
Till now we have found out the local maxima and local minima of the functions but there is one condition on the domain on the function that it should be in the open interval i.e., the end values are not considered.
If the function $f(x)$ is continuous and defined on a closed interval $[a,b]$, then apart from the critical points, we need to find the value of the function at the end-points also. Consider that we have to decide the maximum and minimum values. These maximum and minimum values are called the Absolute/Global Maxima and Absolute/Global Minima respectively. The following steps should be followed to find the absolute maxima and minima of the given function:
- Find all the critical points of $f$ in the given interval.
- Calculate the values of $f(x)$ at all the critical points as well as end points.
- Identify the maximum and minimum values of $f$. The maximum value is the absolute/global maximum value of $f$ and the minimum value is the absolute/global minimum values of $f$.
Question 15. Find absolute maximum and minimum values of a function $f$ given by $$f(x) = 12{x^{\frac{4}{3}}} - 6{x^{\frac{1}{3}}}\;;\;x \in [ - 1,1]$$
Solution: Differentiating the function with respect to $x$, we get $$\begin{equation} \begin{aligned} f(x) = 12{x^{\frac{4}{3}}} - 6{x^{\frac{1}{3}}} \\ f'(x) = 12.\frac{4}{3}{x^{\frac{4}{3} - 1}} - 6.\frac{1}{3}{x^{\frac{1}{3} - 1}} \\ f'(x) = 16{x^{\frac{1}{3}}} - 2{x^{ - \frac{2}{3}}} = \frac{{2(8x - 1)}}{{{x^{\frac{2}{3}}}}} = 0 \\ \Rightarrow x = \frac{1}{8} \\\end{aligned} \end{equation} $$ Also $f'(x)$ is not defined at $x=0$. So, the critical points are $x = 0\;{\text{and }}x = \frac{1}{8}$. Now, we find the value of the function at the critical points as well as end points $x = - 1\;{\text{and }}x = 1$, we get $$\begin{equation} \begin{aligned} f(0) = {12.0^{\frac{4}{3}}} - {6.0^{\frac{1}{3}}} = 0 \\ f\left( {\frac{1}{8}} \right) = 12{\left( {\frac{1}{8}} \right)^{\frac{4}{3}}} - 6{\left( {\frac{1}{8}} \right)^{\frac{1}{3}}} = \frac{{ - 9}}{4} \\ f(1) = 12{\left( 1 \right)^{\frac{4}{3}}} - 6{\left( 1 \right)^{\frac{1}{3}}} = 6 \\ f( - 1) = 12{\left( { - 1} \right)^{\frac{4}{3}}} - 6{\left( { - 1} \right)^{\frac{1}{3}}} = 18 \\\end{aligned} \end{equation} $$ Therefore, the absolute maximum value of $f(x)$ is $18$ that occurs at $x=-1$ and absolute minimum value of $f(x)$ is $\frac{{ - 9}}{4}$ that occurs at ${x = \frac{1}{8}}$.
