Question

Find the range of $f\left( x \right)=\sin x+\cos x+3$. Range is $\left[ a,b \right]$, then find $ab$.

Answer 1

Hint: We will find the maxima and minima of the given function. To find the maxima and minima, we will differentiate the function and equate it to zero. The range of the function will be the values that lie between the minimum value and the maximum value of the function. After we find the range, we will multiply the values of the endpoints of the range to get the desired answer.

Complete step by step solution:
The given function is $f\left( x \right)=\sin x+\cos x+3$. We will look at the derivative of the given function. To find the derivative of a function which is of the form $g\left( x \right)=u\left( x \right)+v\left( x \right)$, we will use the following rule,
$\begin{align}
  & {g}'\left( x \right)=\dfrac{d}{dx}\left( u\left( x \right)+v\left( x \right) \right) \\
 & ={u}'\left( x \right)+{v}'\left( x \right)
\end{align}$
So, we have
$\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( \sin x+\cos x+3 \right)$
Using the rule mentioned above, we get the following,
$\dfrac{d}{dx}\left( \sin x+\cos x+3 \right)=\dfrac{d}{dx}\left( \sin x \right)+\dfrac{d}{dx}\left( \cos x \right)+\dfrac{d}{dx}\left( 3 \right)$
We know that the derivative of a constant function is zero. Therefore, we have $\dfrac{d}{dx}\left( 3 \right)=0$. We also know that $\dfrac{d}{dx}\left( \sin x \right)=\cos x$ and $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$. Substituting these values in the equation above, we get
$\begin{align}
  & \dfrac{d}{dx}\left( f\left( x \right) \right)=\cos x-\sin x+0 \\
 & =\cos x-\sin x
\end{align}$
Now, to find the maxima and minima, we will equate the derivative of the function to zero, that is, ${f}'\left( x \right)=0$. Therefore, we obtain the following equation,
$\cos x-\sin x=0$
This implies that $\cos x=\sin x$. This is possible when $x=\dfrac{\pi }{4}$. The value of the function at $x=\dfrac{\pi }{4}$ is the following,
$f\left( \dfrac{\pi }{4} \right)=\sin \dfrac{\pi }{4}+\cos \dfrac{\pi }{4}+3$
Substituting $\sin \dfrac{\pi }{4}=\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$ in the above equation, we get
$\begin{align}
  & f\left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}+3 \\
 & =\dfrac{2}{\sqrt{2}}+3 \\
 & =\sqrt{2}+3
\end{align}$
We get $\cos x=\sin x$ when $x=\dfrac{\pi }{4}+\pi $, since we know that $\sin \left( \pi +\theta \right)=-\sin \theta $ and $\cos \left( \pi +\theta \right)=-\cos \theta $. Therefore, at $x=\dfrac{\pi }{4}+\pi $, we get the value of the function as
$f\left( \dfrac{\pi }{4}+\pi \right)=\sin \left( \dfrac{\pi }{4}+\pi \right)+\cos \left( \dfrac{\pi }{4}+\pi \right)+3$
Using the identities mentioned above, we get
$\begin{align}
  & f\left( \dfrac{\pi }{4}+\pi \right)=-\sin \dfrac{\pi }{4}-\cos \dfrac{\pi }{4}+3 \\
 & =-\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}+3 \\
 & =-\dfrac{2}{\sqrt{2}}+3 \\
 & =-\sqrt{2}+3
\end{align}$
Hence, the minima of the function is attained at $x=\dfrac{\pi }{4}+\pi $ and the maxima is attained at $x=\dfrac{\pi }{4}$, and their respective values are $-\sqrt{2}+3$ and $\sqrt{2}+3$. So, the range of the given function is $\left[ a,b \right]=\left[ -\sqrt{2}+3,\sqrt{2}+3 \right]$. So, the value of $ab$ can be calculated as follows,
$\begin{align}
  & ab=\left( -\sqrt{2}+3 \right)\left( \sqrt{2}+3 \right) \\
 & =-2-\sqrt{2}\cdot 3+\sqrt{2}\cdot 3+9 \\
 & =7
\end{align}$
Hence, we have $ab=7$.

Note: The maxima and minima of a function are obtained by equating the derivative of the function to zero. The values of $x$ for which we obtain ${f}'\left( x \right)=0$ are called the critical values. We can talk about the function increasing or decreasing by taking the second derivative of the function. It is called the second derivative test.