Question

How do you find the maximum values for $f(x) = \sin x + \cos x$ ?

Answer 1

Hint: For solving this particular problem we need to differentiate the given function with respect to the independent variable that is $x$, then equate the required result to zero. By Solving the equation, we get the values for $x$. Substitute the values in the second derivative of the given function. If the result after substitution is less than zero . Then you have to consider it as the maximum point, and if you get the result greater than zero, you have to consider it as minimum.

Complete step by step solution:
We have $f(x) = \sin x + \cos x$, (given) we need to differentiate the given function with respect to the independent variable that is $x$, then equate the required result to zero. By Solving the equation, we get the values for $x$.
Therefore, we have to differentiate the given function first ,
We will get ,
$f'(x) = \cos x - \sin x$
Now , set $f'(x) = 0$ ,
We will the following result ,
\[ \Rightarrow \cos x - \sin x = 0\]
Add $\sin x$both the side of the equation, we will get ,
\[ \Rightarrow \cos x = \sin x\]
$\Rightarrow x = \dfrac{\pi }{4},\dfrac{{5\pi }}{4}$ in the interval $[0,2\pi ]$ .
The maximum value is ,
$\Rightarrow f(x) = \sin x + \cos x$
$\Rightarrow f\left( {\dfrac{\pi }{4}} \right) = \sin \dfrac{\pi }{4} + \cos \dfrac{\pi }{4}$
$\Rightarrow \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow \sqrt 2 $

Therefore, $\sqrt 2 $ is the maximum value.

Note: A function $f(x)$encompasses a local maximum or relative maximum at $x$ equals to ${x_0}$ if the graph of $f(x)$near ${x_0}$ features a peak at ${x_0}$. A function $f(x)$ features a local minimum or relative minimum at $x$ equals to ${x_0}$ if the graph of $f(x)$ near ${x_0}$ encompasses a trough at ${x_0}$. (To make the excellence clear, sometimes the ‘plain’ maximum and minimum are called absolute maximum and minimum.).