Question

f(x)=sin2x on [0,pi/2]

Answer 1

f'(x) = 2cos2x. f'(x)=2 when x=0
y-pi/2=2(x-0) -using formula (y-y'=m(x-x'))
Y=2x+pi/2 Assuming that you are find the equation for the tangent of the fn f(x) =sin2x

Answer 2

based on rolle's thm...
2cos(2x)=0
cos(2x)=0 my interval is [0,pi/2]
answer u will get as pi/4.

Answer 3

aryande can you explain to me what you are actually doing. I thought we were looking for the tangent of that function. Never heard of this rolle's thm

Answer 4

2cos(2x)=0
cos(2x)=0--->angle is pi/2
for pi/2, cos(2x) = cos(2(pi/2)) = cos(pi) = -1 and yeah cos(2x) = 0 when 2x = pi/2
so by that you can get the interval.

Answer 5

Okay thanks for the explanation.