Question

How do you use the closed interval method to find the absolute maximum and minimum values of the function $f\left( x \right)=x-2\sin x$ on the interval $\left[ -\dfrac{\pi }{4},\dfrac{\pi }{2} \right]$ ?

Answer 1

Hint: For the given question where we need to find the maxima and minima firstly, we need to find the derivative of the function and equate it to zero and then we need to find the local minima and maxima and then compare all values and give the maxima and minima.

Complete step-by-step solution:
In the given question we need to find the absolute maximum and minimum values of the function $f\left( x \right)=x-2\sin x$. So, for this we first need to check whether the given function is continuous or not and we can clearly see that polynomial and trigonometric functions are continuous. Therefore, their difference is also continuous and hence the given function is continuous and we have the closed bounded interval, therefore the function will have the maxima and minima.
So, now we will firstly check the local minima and maxima of the function.
So, for that $f'\left( x \right)=1-2\cos x$
Now,
$\begin{align}
  & f'\left( x \right)\ge 0\Rightarrow 1-2\cos x\ge 0 \\
 & \Rightarrow 2\cos x\le 1 \\
 & \Rightarrow \cos x \le \dfrac{1}{2} \\
 & \Rightarrow x\ge \dfrac{\pi }{3} \\
\end{align}$
Therefore, we get that the function grows in $\dfrac{\pi }{3}\le x\le \dfrac{\pi }{2}$ and declines in $-\dfrac{\pi }{4}\le x\le \dfrac{\pi }{3}$ .
So, now we can see that f’(x) is zero only at one point of the domain which is $\dfrac{\pi }{3}$ .
Therefore, we need to compare values at these three points now
$\dfrac{\pi }{3}$,$-\dfrac{\pi }{4}$,$\dfrac{\pi }{2}$.
$f\left( \dfrac{\pi }{3} \right)=-0.684$
$f\left( -\dfrac{\pi }{4} \right)=0.628$
$f\left( \dfrac{\pi }{2} \right)= -0.43$
And now we clearly say that the function has global minima at $f\left( \dfrac{\pi }{3} \right)$ and maxima at $f\left( -\dfrac{\pi }{4} \right)$.

Note: In the given question, we need to remember that we need to be careful in order to find the roots as sometimes we find the second derivative and equate it to zero and then find the values and compare them to get the answers.