where is f' zero or undefined?
f' = 4secθtanθ + 2sec^2 θ
= 2secθ(2tanθ+secθ)
so, critical numbers are where
secθ = 0 or is undefined,
2tanθ+secθ = 0 or is undefined
That help?
f(theta)= 4sec(theta)+ 2tan(theta), 0 less than theta less than 2pi.
find all critical numbers
6 answers
yes,i got 7pi/6 and 11pi/6 is that correct?
f(Ø) = 4secØ + 2tanØ
for Ø intercepts, f(Ø) = 0
4secØ + 2tanØ = 0
2/cosØ + sinØ/cosØ = 0
times cosØ
2 + sinØ = 0
sinØ = -2 , which is not possible
so the curve does not cross or touch the Ø axis
for y-intercept, let Ø = 0
f(0) = 4sec 0 + 2 tan 0 = 4 + 0 = 4
crosses the vertical axis at (0,4)
f ' (x) = 4 secØtanØ + 2sec^2 Ø
= 0 for max/min of f(Ø)
4/cosØ (sinØ/cosØ) = -2/cos^2 Ø
4 sinØ = -2
sinØ = -1/2
Ø = 7π/6 or Ø = 11π/6
if Ø = 7π/6 , f(7π/6) = 4(-√3/2) + 2/√3 = ....
if Ø = 11π/6, f(11π/6) = ...
I will let you finish the arithmetic
If you want the points of inflection, take the 2nd derivative, and set that equal to zero.
That will be quite a challenge.
for Ø intercepts, f(Ø) = 0
4secØ + 2tanØ = 0
2/cosØ + sinØ/cosØ = 0
times cosØ
2 + sinØ = 0
sinØ = -2 , which is not possible
so the curve does not cross or touch the Ø axis
for y-intercept, let Ø = 0
f(0) = 4sec 0 + 2 tan 0 = 4 + 0 = 4
crosses the vertical axis at (0,4)
f ' (x) = 4 secØtanØ + 2sec^2 Ø
= 0 for max/min of f(Ø)
4/cosØ (sinØ/cosØ) = -2/cos^2 Ø
4 sinØ = -2
sinØ = -1/2
Ø = 7π/6 or Ø = 11π/6
if Ø = 7π/6 , f(7π/6) = 4(-√3/2) + 2/√3 = ....
if Ø = 11π/6, f(11π/6) = ...
I will let you finish the arithmetic
If you want the points of inflection, take the 2nd derivative, and set that equal to zero.
That will be quite a challenge.
Thank you!!
Hmm. I was under the impression that critical numbers were numbers in the domain where f'=0 or f is not differentiable.
I think your math is great, Reiny, but maybe covers too much ground.
I think your math is great, Reiny, but maybe covers too much ground.
f (0)=sinø + cos ø of(0)=3 f'(0)=4
6 answers