• Support PF! Buy your school textbooks, materials and every day products Here!

Investigating max and min value of a function

  • #1
1. Homework Statement

[tex] f(x) = (1/2)sin2x + cosx [/tex]

Find

[tex] f^2 min +f^2 max = ? [/tex]



2. Homework Equations

Differentiation not allowed... only by transformations and analysis.

3. The Attempt at a Solution

I am confused by what it means by f^2 min +f^2 max... does it imply we have to find max and min values seperately, square them and add them? Or does this formulation imply we can directly get the required value?

Do we have to square the function before investigating it?
 

Answers and Replies

  • #2
dynamicsolo
Homework Helper
1,648
4
[tex] f(x) = (1/2)sin2x + cosx [/tex]
I think what they're asking you to do is find the sum of the squares of the maximum and minimum values of this function.

I'd suggest first writing out what sin(2x) is: you'll find that f(x) can be expressed as a product of two factors. It should be reasonably straightforward to see what the greatest and least values of that product are. Can f(x) be zero?
 
  • #3
1,395
0
you need to take the derivative of that..

[tex] f(x) = (1/2)sin2x + cosx [/tex]


f'(x)=1/4*cos2x-sinx
after that you solve f'(x)=0
find the extreme points and put the y values in the asked equation

[tex] f^2 min +f^2 max = ? [/tex]
 
  • #4
dynamicsolo
Homework Helper
1,648
4
[tex] f(x) = (1/2)sin2x + cosx [/tex]


f'(x)=1/4*cos2x-sinx
after that you solve f'(x)=0
find the extreme points and put the y values in the asked equation

[tex] f^2 min +f^2 max = ? [/tex]
The problem statement says no differentiation allowed...

(If it were, the derivative would be f'(x)=1*cos2x - sinx .)
 
Last edited:
  • #5
i'm getting

[tex]f(x) = cosx(1+sinx)[/tex]

ok i got min value = 0 (the problem says 0 <= x <= pi/2)
how do we get the max value of the product?
 
  • #6
1,395
0
you have a point in x=pi/2 +pi*k
and for x=3/4*pi +2pi*k

substitute them in the fuction and find their y values
 
Last edited:
  • #7
how does that work?
 
  • #8
dynamicsolo
Homework Helper
1,648
4
how does that work?
If the interval is [0, pi/2], then you don't have to worry about the other periodic values of sine and cosine. (You didn't mention the interval earlier...)

The minimum is zero at pi/2 because of the cosine term. For the maximum, you could either look at the terms in f(x) or square your result for f(x) first. In any case, using x = 0 would give you
f(0) = 1, but there's a place where we can do better. The problem with the endpoints is that sine is high when cosine is low and vice versa. What value of x gives both fairly high values for sine and cosine? (Consider graphs of those functions.)
 
  • #9
pi/4 gives equal values for sine and cosine... however, how do we know there isn't a value thats higher? i think there may be a more rigorous proof...

e.g. on another similar problem i could obtain a quadratic equation containing f(x) in a constant term (by squaring function) and getting something like:

[tex] ax^2 +bx + c = f^2(x) [/tex]

[tex] ax^2 +bx + (c - f^2(x)) = 0 [/tex]

[tex] b^2 - 4a(c - f^2(x)) >or= 0 [/tex] (for real f(x))

thus obtaining min and max values simultaneously...

on this example, on squaring i get a quartic equation on sin(x)... i'm unable to bring it to a simpler (quadratic) form or otherwise...
 
Last edited:
  • #10
dynamicsolo
Homework Helper
1,648
4
Maybe we shouldn't look at the quartic polynomial, but rather the factored form. The function squared is

[tex] (cos x)^{2} (1+sin x)^{2} = (1 - [sin x]^{2})(1+sin x)^{2} = (1 - sin x)(1+sin x)^{3}
[/tex].

So we make the substitution t = sin x and ask for the maximum on the interval [0,1] of

[tex] (1 - t)(1+ t)^{3}
[/tex].

[I'm still thinking about how to solve this without calculus. (The maximum turns out to occur at sin x = 1/2 , BTW, not {sqrt(2)}/2 , as I'd earlier thought.) The instruction "use transformations and analysis" isn't very descriptive, so I'm still trying out ideas...]
 
Last edited:

Related Threads for: Investigating max and min value of a function

Replies
1
Views
4K
Replies
2
Views
704
  • Last Post
Replies
13
Views
348
  • Last Post
Replies
3
Views
821
  • Last Post
Replies
6
Views
1K
Replies
11
Views
842
Replies
6
Views
3K
  • Last Post
Replies
7
Views
2K
Top