URGENT:1) show that the function f(x)= tanx-4x is decreasing function on (-pie/3,pie/3). 2) Show that f(x)=x2-x sinx is an increasing function on (0,pie/2)
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f(x ) = tanx -4x
f'(x ) = sec²x -4
f'(x) =(secx -2)(secx +2)
in interval (-π/3, π/3)
secx€ ( 1, 2)
so, f'(x) <0
function is decreasing only when
f'(x )<0
hence function is decreasing.
f(x) =x² -xsinx
f'(x) =2x -sinx +xcosx
not clear so, again differentiate
f"( x) = 2 -cosx + cosx -xsinx
f"(x) = 2 -xsinx
0 < x < π/2
0 < sinx < 1
both are increasing in( 0,π/2) so, multiple of both will not change the sign
0 < xsinx< π/2
-π/2 < -xsinx<0
-π/2 +2 <2 -xsinx <2
here we see f"(x)>0
so, function f(x) is increasing
f'(x ) = sec²x -4
f'(x) =(secx -2)(secx +2)
in interval (-π/3, π/3)
secx€ ( 1, 2)
so, f'(x) <0
function is decreasing only when
f'(x )<0
hence function is decreasing.
f(x) =x² -xsinx
f'(x) =2x -sinx +xcosx
not clear so, again differentiate
f"( x) = 2 -cosx + cosx -xsinx
f"(x) = 2 -xsinx
0 < x < π/2
0 < sinx < 1
both are increasing in( 0,π/2) so, multiple of both will not change the sign
0 < xsinx< π/2
-π/2 < -xsinx<0
-π/2 +2 <2 -xsinx <2
here we see f"(x)>0
so, function f(x) is increasing
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