Question

Find the maximum and minimum values of the following expression for 0<=theta<360.

\(\frac{ 1 }{ 6+\cos ^{2} \Theta }\)

@Callisto

Answer 1

The smaller the denominator, the larger the value of the function, agree?

Answer 2

i don't understand actually

Answer 3

1/2 and 1/4, which one is larger?

Answer 4

1/2

Answer 5

ok, got it

Answer 6

So, to find the max, we need to minimize the denominator.

Consider the range of cosθ, -1 <= cosθ <= 1.
So, what is the range of (cosθ)^2?

Answer 7

1<=cosθ<=1
???!!

Answer 8

let \[\frac{1}{6+\cos^2\theta}=f(\theta)\]
\[f'(\theta)=\frac{\sin2\theta}{(6+\cos^2\theta )^2}=0\implies \sin 2\theta =0,2\theta=90+360k\\ \therefore \theta=45+180k,\theta=45,225\\f(45)=2/13=f(225)\]

Answer 9

1<=cosθ<=1
^That means cosθ = 1...
It should be 0<= cosθ <=1.
Understand why?

Answer 10

1/6

Answer 11

not sure about min

Answer 12

@Callisto understand

Answer 13

So, you can find the min and max now.
Substitute the min of (cosθ)^2 to find the max of the function; and substitute the max of (cosθ)^2 to find the min of the function.

Answer 14

1/7 <= ... <= 1/6

Answer 15

So, do you get the answer now?

Answer 16

max : 1/6
min : 1/7

Answer 17

Yup :)

Answer 18

thanks ;D
i will remember the rule