Question

Find the points of local maxima or local minima of the following function, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be.
$f(x) = \sin 2x$,$0 < x < \pi$

Answer 1

For calculating maximum or minimum value of any function, we find the first derivation of a given function. After putting the first derivative equal to zero, we get points at which function is maximum or minimum. We check the values for critical points.

Formula used: In question we differentiate the function $f(x) = \sin 2x$ with respect to x. Only differentiation methods are used.

Complete step-by-step answer:
Given: $f(x) = \sin 2x$
On differentiating function$f(x)$with respect to x , we get
$\dfrac{d}{{dx}}f(x) = \dfrac{d}{{dx}}\sin 2x$
\Rightarrow \dfrac{d}{{dx}}f(x) = 2\cos 2x$$$$\left[ {\because \dfrac{d}{{dx}}(\sin ax) = a\cos ax} \right]
${f'}(x = 2\cos 2x$ $\left[ {\because \dfrac{d}{{dx}}f(x) = {f'}(x} \right]$
For the critical point, ${f'}(x) = 0$
$2\cos 2x = 0$
$\Rightarrow \cos 2x = 0$
$\Rightarrow \cos x = 0$
$\Rightarrow 2x = \dfrac{\pi }{2},\dfrac{{3\pi }}{2}$
$\therefore x = \dfrac{\pi }{4},\dfrac{{3\pi }}{4}$
Hence, critical points are $\dfrac{\pi }{4}$ and $\dfrac{{3\pi }}{4}$
For first critical point,$x = \dfrac{\pi }{4}$
${f'}(x)$ Value changes from positive to negative.
Hence $x = \dfrac{\pi }{4}$ is known as point of maxima,
So value of $f(x)$ at $x = \dfrac{\pi }{4}$ is given by
$f\left( {\dfrac{\pi }{4}} \right) = \sin 2\left( {\dfrac{\pi }{4}} \right)$
$\Rightarrow f\left( {\dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{\pi }{2}} \right)$
$\therefore f\left( {\dfrac{\pi }{4}} \right) = 1$
For second point,$x = \dfrac{{3\pi }}{4}$
$f’(x)$ value changes from negative to positive.
Hence $x = \dfrac{{3\pi }}{4}$ is known as point of minima,
So value of $f(x)$ at $x = \dfrac{{3\pi }}{4}$ is given by
$f\left( {\dfrac{{3\pi }}{4}} \right) = \sin 2\left( {\dfrac{{3\pi }}{4}} \right)$
$\Rightarrow f\left( {\dfrac{\pi }{4}} \right) = \sin \left( {\dfrac{{3\pi }}{2}} \right)$
$\therefore f\left( {\dfrac{\pi }{4}} \right) = - 1$

Hence the local maximum value of $f(x) = \sin 2x$ is $1$ and local minimum value is $- 1$.

Note: The differentiation of trigonometric functions is defined as the mathematical process, in which the derivative of a trigonometric function is found. In other hand it may be described as the rate of change with respect to a variable.
The maxima and minima of a function are also known as the largest and smallest value. The respective plurals of maximum and minimum are called maxima and minima. In question, terms local maximum and local minimum are used. A local maximum at ${\text{x }} = {\text{ a}}$ of a function $f(x)$ is described as, if The value of $f(x)$ at $x = a$ is more than the value of $f(x)$
i.e, $f(a) > f(x)$
Similarly, A local minimum of a function $f(x)$ is given as, at ${\text{x }} = {\text{ a}}$, if The value of $f(x)$ at ${\text{x = a}}$ is less than the value of $f(x)$
i.e$f(a) < f(x)$