question - help appreciated (2 Viewers)

[lyounamu]

Find the maximum and minimum values of cosX/(1+sinX), 0 <= x <= pi

This is a 5-mark question so I just couldn't use my common sense to answer this question.

I found the derivative but X works out to be 3pi/2 which is out of the range and this cannot be the solution as well beucase 1+sin3pi/2 = 0 which makes the denominator 0, hence no solution.

I tried to explain that the maximum value of cosX/(1+sinX) = 1 because
the maximum point occurs when cosX = 1 and sinX = 0, therefore X = 0

And the minimum value of cosX/(1+sinX) = -1 because the minimum points occurs when cosX = -1 and sin X= 0, therefore X=pi

But this is a bloody 5-mark question so I need some decent working out. Any help will be appreciated. Thanks!

 

[lolokay]

if there aren't any min or maxes in the range, then it must be the case that the min and max within the range occur at the extremities, ie 0 and pi
cos0/(1+sin0) = 1
cospi/(1+sinpi) = -1

so they are the max and min values

I don't know how you can get 5 marks out of that question though

 

[lyounamu]

if there aren't any min or maxes in the range, then it must be the case that the min and max within the range occur at the extremities, ie 0 and pi
cos0/(1+sin0) = 1
cospi/(1+sinpi) = -1

so they are the max and min values

I don't know how you can get 5 marks out of that question though

May be I was expected to draw graph or something. (even though question doesn't specify that)

Well, I would have drawn graph and done the same thing (as I and you already did) anyway if I couldn't use the derivative...lol

Thanks for confirming that.

 

[Trebla]

This question is basically acknowledging that the absolute/global maximum or minimum is not the same as the local maximum or minimum.

I think the 5 marks allocated are:
- find the derivative and recognise that there is no turning point in the given domain
- realise that the function is decreasing in the domain
- acknowledging the min occurs at the upper bound of the domain
- acknowledging that the max occurs in the lower bound of the domain
- finding the actual absolute maximum and minimum values

In your example, y = cos x/(1 + sin x)
dy/dx = [- sin x(1 + sin x) - cos²x] / (1 + sin x)²
= [- sin x - (sin²x + cos²x)] / (1 + sin x)²
= [- sin x - 1]/(1 + sin x)²
Stationary points occur at sin x = -1
However since 0 ≤ x ≤ π, no solution exists (as sin x is non-negative in that domain), so there are no stationary points in the given domain.
Since sin x is non-negative, - sin x - 1 is clearly negative, thus
dy/dx = [- sin x - 1]/(1 + sin x)² < 0 for 0 ≤ x ≤ π
This means the function is monotonically decreasing in that domain (i.e. the larger the x, the smaller the y). The fact that its decreasing (downward sloping) tells you that the maximum occurs at x = 0 (y = 1) and the minimum occurs at x = π (y = -1) without having to sketch the graph.

 

[lyounamu]

This question is basically acknowledging that the absolute/global maximum or minimum is not the same as the local maximum or minimum.

I think the 5 marks allocated are:
- find the derivative and recognise that there is no turning point in the given domain
- realise that the function is decreasing in the domain
- acknowledging the min occurs at the upper bound of the domain
- acknowledging that the max occurs in the lower bound of the domain
- finding the actual absolute maximum and minimum values

In your example, y = cos x/(1 + sin x)
dy/dx = [- sin x(1 + sin x) - cos²x] / (1 + sin x)²
= [- sin x - (sin²x + cos²x)] / (1 + sin x)²
= [- sin x - 1]/(1 + sin x)²
Stationary points occur at sin x = -1
However since 0 ≤ x ≤ π, no solution exists (as sin x is non-negative in that domain), so there are no stationary points in the given domain.
Since sin x is non-negative, - sin x - 1 is clearly negative, thus
dy/dx = [- sin x - 1]/(1 + sin x)² < 0 for 0 ≤ x ≤ π
This means the function is monotonically decreasing in that domain (i.e. the larger the x, the smaller the y). The fact that its decreasing (downward sloping) tells you that the maximum occurs at x = 0 (y = 1) and the minimum occurs at x = π (y = -1) without having to sketch the graph.

Thanks! You are truly a life-saver!