Question

How do you locate the absolute extrema of the function $f\left( x \right)={{x}^{3}}-12x$ on the closed interval $\left[ 0,4 \right]$ ?

Answer 1

Hint: Here in this question we have been asked to locate the absolute extrema of the given function $f\left( x \right)={{x}^{3}}-12x$ in the closed interval $\left[ 0,4 \right]$. We know that the absolute extrema of a function $f$ , that is continuous in a closed interval must occur at either a critical number for $f$ in the interval or at an endpoint of the interval.

Complete step by step answer:
Now considering from the question we have been asked to locate the absolute extrema of the given function $f\left( x \right)={{x}^{3}}-12x$ in the closed interval $\left[ 0,4 \right]$.
From the basic concepts we know that the absolute extrema of a function $f$ , that is continuous in a closed interval must occur at either a critical number for $f$ in the interval or at an endpoint of the interval.
We know that the critical number for $f$ is the zero for $f'$ .
Now we can say that $f'\left( x \right)=3{{x}^{2}}-12$ since $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ .
The zeroes of $f'$ will be generally given as
$\begin{align}
  & f'\left( x \right)=3\left( {{x}^{2}}-4 \right) \\
 & \Rightarrow f'\left( x \right)=3\left( x-2 \right)\left( x+2 \right) \\
\end{align}$ .
Therefore the zeroes of $f'$ are $2,-2$ and in the closed interval $\left[ 0,4 \right]$ we have only one critical point for $f$ that is $2$ and the endpoints will be $0,4$.
Hence we can say that the absolute extrema of the given function $f\left( x \right)={{x}^{3}}-12x$ in the closed interval $\left[ 0,4 \right]$ will be located at $x=0,2,4$ any of these points.
So now we need to evaluate the value of $f\left( x \right)$ at these points.
At $x=0$ , the value of $f\left( x \right)={{x}^{3}}-12x$ will be zero.
At $x=2$ , the value of $f\left( x \right)={{x}^{3}}-12x$ will be $-16$ .
At $x=4$ , the value of $f\left( x \right)={{x}^{3}}-12x$ will be $16$ .

Therefore we can conclude that the absolute extrema of the given function $f\left( x \right)={{x}^{3}}-12x$ in the closed interval $\left[ 0,4 \right]$will be located at $x=2$ (minimum) and $x=4$ (maximum).

Note: While answering this type of questions we have to be sure with the calculations that we are going to perform and the concepts that we are going to apply in between the steps in order to answer them. Alternatively this question can be answered by using the graph of $f\left( x \right)={{x}^{3}}-12x$ which is generally given as: