Question

f(x)= sin(arctanx). find the range of f.

Answer 1

y = sin( arctanx)

we know, tan^-1x or arctanx defined when , -π/2 < x < π/2
hence,
domain of the function is ( -π/2 , π/2)

now, we check function is increasing or decreasing nature .
y = sin(arctanx)
differentiate wrt x
dy/dx = cos(arctanx)× 1/( 1 + x²)

-π/2 < x < π/2
-∞ < arctanx< ∞
hence we can't say cos(arctanx) is positive or negative . it will be both .


hence, function are both nature increasing and decreasing .
so, range will be fined by checking in x = ∞, -∞, π/2 and -π/2

y = f(x) = sin( arctanx)
f( -∞) = sin( arctan(-∞) = sin( -π/2) = -1
f( ∞} = sin( arctan( ∞) = sin(π/2) = 1
f( π/2) = sin( arctanπ/2) = π/√(π² +4)
f( -π/2) =sin( arctan(-π/2)) = -π/√(π² +4)
hence,
maximum value of f(x) = 1
minimum value of f(x) = -1

so, range = ( -1 , 1)