The Student Room Group

c3 differentiation..heeelllpppp!

heyy can someone explain hw to do these questions?

1) find the minimum value of y= tanx-8sinx which is given by a value of x between 0 and pi/2

answer is -3√3

2) show that for x is bigger than 0 but smaller than pi/2, f(x)=xsinx+cosx is an increasing function.

3) find the co-ords of the stationary points on the curve y=sinx + cosx
state which is maximum and minimum.

answer: (pi/4,√2) = maximum, (5pi/4, -√2) = maximum

thanks guys :smile:
Reply 1
any1?
Fizzle
heyy can someone explain hw to do these questions?

1) find the minimum value of y= tanx-8sinx which is given by a value of x between 0 and pi/2

answer is -3√3

2) show that for x is bigger than 0 but smaller than pi/2, f(x)=xsinx+cosx is an increasing function.

3) find the co-ords of the stationary points on the curve y=sinx + cosx
state which is maximum and minimum.

answer: (pi/4,√2) = maximum, (5pi/4, -√2) = maximum

thanks guys :smile:

1) y'=sec^2x-8cos(x)
=1-8cos^3(x)/cos^2x
this equals 0 when
cos^3(x)=1/8
cos(x)=1/2
so x=pi/3
at this value y=rt(3)-8(rt(3)/2)=rt(3)-4rt(3)=-3rt(3).

2) f'(x)=sinx+ xcosx -sinx= xcosx
on the given interval x is non-zero and cosx is non-zero hence f'(X) is non-zero. Thus f(x) has no turning points so is either increasing or decreaing.
f(pi/4)<f(pi/3) and pi/4<pi/3 so it is increasing.


3)
f'(x)=cos x-sinx
this is zero when cos x=sinx ie tanx =1
giving x=pi/4 and x=5pi/4
y=rt(2) y=-rt(2)
f''(x)=-cosx-sinx
f''(pi/4)=-2/(rt(2))<0 so this point is a maximum
f''(5pi/4)=2/(rt(2)>0 so this point is a minimum
Reply 3
cheers, dont understand some points but ill ask my teacher!
Fizzle
cheers, dont understand some points but ill ask my teacher!

what points ? i could try and help.