Question

Find the absolute maximum(maxima) and minimum (minima) values of the function f given by $f(x) = {\cos ^2}x + \sin x,x \in [0,\pi ]$ .

Answer 1

Hint: Here we will use the concept of maxima and minima of a function, known collectively as extrema, are the largest and smallest value of the function, either within a given range or on the entire domain. First we will find the derivative of the function then equate it to zero. Then we will check the value of the function at these critical points which will give absolute maxima and minima.

Complete step by step answer:
${f^{'}}(x)$$ = - 2\cos x\sin x + \cos x$
On equating this value of ${f^{'}}(x)$ to zero, we get
${f^{'}}(x)$$ = - 2\cos x\sin x + \cos x$
$ \Rightarrow - 2\cos x\sin x + \cos x = 0$
$\cos x(2\sin x - 1)=0$
$ \Rightarrow x = \dfrac{\pi }{6}or\dfrac{\pi }{2}$
Now let's evaluate the value of the function at critical points and at extreme points of domain.
$f\left( {\dfrac{\pi }{6}} \right) = {\cos ^2}\left( {\dfrac{\pi }{6}} \right) + \sin \left( {\dfrac{\pi }{6}} \right) = \dfrac{5}{4}$
$f\left( {\dfrac{\pi }{2}} \right) = {\cos ^2}\left( {\dfrac{\pi }{2}} \right) + \sin \left( {\dfrac{\pi }{2}} \right) = 1$
$f(0) = {\cos ^2}(0) + \sin (0) = 1$
$f(\pi ) = {\cos ^2}(\pi ) + \sin (\pi ) = 1$
And we can see that function will have maxima at $x = \dfrac{\pi }{6}$ and will have minima at $x = \dfrac{\pi }{2},0,\pi $

Therefore, we say the absolute maximum(maxima) and minimum (minima) values of the given function are $\dfrac{5}{4}$ and 1 respectively.

Note: we need to know the range of the all trigonometry function on the entire range. As we know Basic functions in trigonometry are sine, cos and tan. The values of sin vary from 0 (for angle of 0 degrees) to 1.0 for 90 degrees. The values of cos vary from 1.0 (for angle of 0 degrees) to 0 for 90 degrees, while the values of tan vary from 0 (for angle of 0 degrees) to infinity for 90 degrees.