Functions. Least/Greatest value and f(x)+1=0 issue.

[SolCon]

Hi. :)

Alright, another functions related problem. This question came in 3 parts. I've done the first one but the other 2 are being a problem. Here it is:

Q. The function f is such that f(x) = 2 sin^2(x) − 3 cos^2(x) for 0 ≤ x ≤ p.

(i) Express f(x) in the form a + b cos^2(x), stating the values of a and b.

This was done easily. Answer came as 2-5cos^2(x) + 1 =0, where a=2, b=-5.

(ii) State the greatest and least values of f(x).

I think we have to make a graph of the function here. Problem is, there's cos^2 instead of just cos. I'm not sure how to do this. If there is some other way to find the problem, I'd like to try that too.

(iii) Solve the equation f(x) + 1 = 0.

Well, I tried this but got stuck. Here's what I did:

2-5cos^2(x) + 1 = 0
2+1 = 5cos^2(x)
3=5cos^2(x)
3/5=cos^2(x)

And then I'm drawn a blank. What to do next or if I've even done it right.

Any help is appreciated. :)

 

[Mark44]

Hi. :)

Alright, another functions related problem. This question came in 3 parts. I've done the first one but the other 2 are being a problem. Here it is:

Q. The function f is such that f(x) = 2 sin^2(x) − 3 cos^2(x) for 0 ≤ x ≤ p.

(i) Express f(x) in the form a + b cos^2(x), stating the values of a and b.

This was done easily. Answer came as 2-5cos^2(x) + 1 =0, where a=2, b=-5.

Your answer is incorrect. For one thing, what you show is really 3 - 5cos²(x) = 0. Where did that +1 you show come from? It shouldn't be there.

For another thing, your answer shouldn't be an equation with 0 on one side. All they're asking you to do is to rewrite the formula for f(x) in a different form.

What you should have gotten was f(x) = 2 - 5cos²(x).

(ii) State the greatest and least values of f(x).

I think we have to make a graph of the function here. Problem is, there's cos^2 instead of just cos. I'm not sure how to do this. If there is some other way to find the problem, I'd like to try that too.

You can get a rough graph of y = cos²x by taking the parts of the graph y = cos(x) that lie below the x-axis and reflecting them across the x-axis. Since all you're interested in are the maximum and minimum values, the inexactness of this technique won't really matter.

Once you have a rough graph of y = cos²(x), you can get a rough sketch of y = 2 - 5cos²(x) by scaling, reflecting, and translating the graph of y = cos²(x).

(iii) Solve the equation f(x) + 1 = 0.

Well, I tried this but got stuck. Here's what I did:

2-5cos^2(x) + 1 = 0
2+1 = 5cos^2(x)
3=5cos^2(x)
3/5=cos^2(x)

Maybe this is where that extra +1 term came from.

Take the square root of both sides, keeping in mind that you should get both a positive and a negative answer.

And then I'm drawn a blank. What to do next or if I've even done it right.

Any help is appreciated. :)

 

[SolCon]

Your answer is incorrect. For one thing, what you show is really 3 - 5cos2(x) = 0. Where did that +1 you show come from? It shouldn't be there.

For another thing, your answer shouldn't be an equation with 0 on one side. All they're asking you to do is to rewrite the formula for f(x) in a different form.

What you should have gotten was f(x) = 2 - 5cos2(x).

This was my mistake.
I read and typed that part of the page where I had written part (i)'s answer and part (iii)'s solution. Yes, that is exactly what I'm getting. :)

You can get a rough graph of y = cos2x by taking the parts of the graph y = cos(x) that lie below the x-axis and reflecting them across the x-axis. Since all you're interested in are the maximum and minimum values, the inexactness of this technique won't really matter.

Once you have a rough graph of y = cos2(x), you can get a rough sketch of y = 2 - 5cos2(x) by scaling, reflecting, and translating the graph of y = cos2(x).

Right, but what values do I have to take for x? Do I have to take, for example, cos (pi/2), cos (pi), cos (3pi/2), cos (2pi) etc then square them for y=cos^2(x)?

Maybe this is where that extra +1 term came from.

Take the square root of both sides, keeping in mind that you should get both a positive and a negative answer.

Thanks. Got the answer now. :)