Question

The maximum value and minimum value, if any, of the function f (x)=sin² x, are given as ______ and ______, respectively.​

Answer 1

Answer:

max = 1, and min = 0

Step-by-step explanation:

f(x) = sin²x

f'(x) = 2sinx.cosx = sin2x

f"(x) = 2cos2x

for , maximum and minimum

f'(x) = 0 , sin2x =0 , 2x = 0, π , x = 0, π/2

Now,

x = 0, f"(0) = 2cos0 = 2>0

f(x) is minimum at x = 0

minimum value = sin²0 = 0.

x = π/2 , f"(π/2) = 2cosπ = -2<0

f(x) is maximum at x = π/2

maximum value = sin²π/2 = 1.

Answer 2

Given: f(x) = sin² x

To find: The maximum value and minimum value of the function.

Solution:

To find the maximum value and minimum value of the function, the first and the second derivative of the given function is to be found.

The first derivative is calculated as

$f'(x) = 2 (sin x) (cos x)$

        $= sin 2x$

The second derivative is calculated as

$f''(x) = 2 (cos2x)$

The first derivative is equated to zero and the two values of x are found. These values are substituted in place of x in the second derivative. If the answer obtained is greater than zero, then f(x) is minimum for that value of x and if the answer obtained is lesser than zero, then f(x) is maximum for that value of x. Finally, the maximum value and minimum value of the function for the obtained x value are calculated.

$sin 2x = 0$

$x = 0, \frac{\pi }{2}$

For x=0,

$f''(0) = 2 cos(0)$

         $= 2 >0$

$sin 0 = 0$

For x = π/2,

$f''(\frac{\pi }{2}) = 2 cos \pi$

          $= -2 < 0$

$sin \frac{\pi }{2} = 1$

Therefore, the maximum value and minimum value of the function are 0 and 1, respectively.