f(x)=Sin²x x ∈ [0, 2π],Find the maximum and minimum value of the given function.
Answers
Answered by
2
Dear Student,
Solution:
1) Differentiate given function with respect to x
2) put f'(x) = 0,find the values of x
3) differentiate f'(x) again to obtain f''(x)
4) Check for that values of x, if f''(x) > 0, function has minima at that point and if
f''(x) <0, function has maxima at that point.
f(x)=Sin²x x ∈ [0, 2π],
f'(x) = 2 sin x cos x
= sin 2x
put f'(x) = 0
sin 2x = 0
sin 2x = sin ( n)π n are integer ∈ R ,
case 1: n =0
2x = 0
x = 0
case 2: n = 1
2x = π
x = π/2
case 3: n = 2
2x = 2π
x = π
f''(x) = 2 cos 2x
put x =0, f"(x) = 2 cos 0 = 2 >0
at x = 0, there is a Maxima
at x = π/2
f"(x) = 2 cos π = -2 <0 ,maxima at x = π/2
By this way maxima and minima of a function can be calculate.
Hope it helps you
Solution:
1) Differentiate given function with respect to x
2) put f'(x) = 0,find the values of x
3) differentiate f'(x) again to obtain f''(x)
4) Check for that values of x, if f''(x) > 0, function has minima at that point and if
f''(x) <0, function has maxima at that point.
f(x)=Sin²x x ∈ [0, 2π],
f'(x) = 2 sin x cos x
= sin 2x
put f'(x) = 0
sin 2x = 0
sin 2x = sin ( n)π n are integer ∈ R ,
case 1: n =0
2x = 0
x = 0
case 2: n = 1
2x = π
x = π/2
case 3: n = 2
2x = 2π
x = π
f''(x) = 2 cos 2x
put x =0, f"(x) = 2 cos 0 = 2 >0
at x = 0, there is a Maxima
at x = π/2
f"(x) = 2 cos π = -2 <0 ,maxima at x = π/2
By this way maxima and minima of a function can be calculate.
Hope it helps you
abhi178:
sir, for for disturbing . but question says find absolute maximum and absolute minimum indirectly because they want maximum value of f(x) and minimum value of f(x) too
that is ,I am writing,Maximum value 1
Answered by
2
it is given that f(x) = sin²x ,x ∈ [0, 2π]
we havr to find maximum and minimun value of function in interval [0, 2π]
first, Let's find critical points :
f(x) = sin²x
differentiate with respect to x,
f'(x) = 2sinx. cosx
now, f'(x) = 0
so, 2sinx.cosx = 0
sinx = 0 => x = 0, π , 2π [because x belongs to [0, 2π]
cosx = 0 => x = π/2, 3π/2 [ because x belongs to [0, 2π ]
hence, critical points are : x = 0, π/2, π, 3π/2, 2π.
now, find value of f(x) at x = 0, π/2, π , 3π/2 , 2π
f(0) = sin²0 = 0
f(π/2) = sin²π/2 = 1
f(π) = sin²π = 0
f(3π/2) = sin²3π/2 = (-1)² = 1
f(2π) = sin²π = 0
here it is clear that f(x) attains maximum value = 1 and minimum value 0.
we havr to find maximum and minimun value of function in interval [0, 2π]
first, Let's find critical points :
f(x) = sin²x
differentiate with respect to x,
f'(x) = 2sinx. cosx
now, f'(x) = 0
so, 2sinx.cosx = 0
sinx = 0 => x = 0, π , 2π [because x belongs to [0, 2π]
cosx = 0 => x = π/2, 3π/2 [ because x belongs to [0, 2π ]
hence, critical points are : x = 0, π/2, π, 3π/2, 2π.
now, find value of f(x) at x = 0, π/2, π , 3π/2 , 2π
f(0) = sin²0 = 0
f(π/2) = sin²π/2 = 1
f(π) = sin²π = 0
f(3π/2) = sin²3π/2 = (-1)² = 1
f(2π) = sin²π = 0
here it is clear that f(x) attains maximum value = 1 and minimum value 0.
Similar questions