Max/Min of f(x) = cos 2x + 2sinx for 0 ≤ x ≤ 3π/4

[fghtffyrdmns]

Homework Statement

Find the maximum value and minimum value of the function f(x) = cos 2x + 2sinx

0 \leq x \leq \frac{3\pi}{4}

Homework Equations

f(x) = cos 2x + 2sinx

The Attempt at a Solution

f(x) = cos 2x + 2sinx

f'(x) = -2sin2x + 2cosx

0=-2sin2x + 2 cosx

2sin2x=2cosx

sin2x = cosx

2sinxcosx = cosx

2sinx = 1

sin x = \frac{1}{2}

x = \frac{\pi}{6}

f''(x) = -4cos2x - 2sinx
If I check, it is a max. But everything(in the domain) I check is max. What's up with this? Is my derivative wrong?

 

[slider142]

You have several values you did not check. You must check the endpoints of the interval, x=0 and x=\frac{3\pi}{4}. they may or may not be turning points, but their values may yet be higher or lower than any turning point in the interior of the interval.
You also divided by cos(x) in one step, which means you must go back and see what happens to f'(x) when cos(x) = 0, as your solution from that point onwards assumes that cos(x) is not 0. First, check to see whether cos(x) is ever 0 in the interval given. If so, you must check that point(s) as well. This, in fact, gives you an additional critical point where sin(2x) = 0.
Your derivatives are correct. Why are you checking every point in the domain? Remember, the second derivative only gives you concavity/curvature. It does not, by itself, tell you whether a point is a maximum or minimum unless that point is also a critical point.

 

[Mark44]

Homework Statement

Find the maximum value and minimum value of the function f(x) = cos 2x + 2sinx

0 \leq x \leq \frac{3\pi}{4}

Homework Equations

f(x) = cos 2x + 2sinx

The Attempt at a Solution

f(x) = cos 2x + 2sinx

f'(x) = -2sin2x + 2cosx

0=-2sin2x + 2 cosx

2sin2x=2cosx

sin2x = cosx

2sinxcosx = cosx

In the line below, you have divided both sides by cosx, effectively discarding cos x = 0 as a solution to your original equation.

You should have changed the equation above to this:
2sinxcosx - cos x = 0
==> cosx(2sinx - 1) = 0
==> cos x = 0 or sinx = 1/2

2sinx = 1

sin x = \frac{1}{2}

x = \frac{\pi}{6}

f''(x) = -4cos2x - 2sinx
If I check, it is a max. But everything(in the domain) I check is max. What's up with this? Is my derivative wrong?

I don't understand what you're saying when everything in the domain is a max. On the restricted domain there are a local minimum and a local maximum. Since the domain is restricted, you should be able to find an absolute min and absolute max.

 

[fghtffyrdmns]

You have several values you did not check. You must check the endpoints of the interval, x=0 and x=\frac{3\pi}{4}. they may or may not be turning points, but their values may yet be higher or lower than any turning point in the interior of the interval.
You also divided by cos(x) in one step, which means you must go back and see what happens to f'(x) when cos(x) = 0, as your solution from that point onwards assumes that cos(x) is not 0. First, check to see whether cos(x) is ever 0 in the interval given. If so, you must check that point(s) as well. This, in fact, gives you an additional critical point where sin(2x) = 0.
Your derivatives are correct. Why are you checking every point in the domain? Remember, the second derivative only gives you concavity/curvature. It does not, by itself, tell you whether a point is a maximum or minimum unless that point is also a critical point.

Ah, I should've worded it better. I checked x=\frac{\pi}{6} with the second derivative. I got -3 which makes it a local max.

 

[slider142]

Ah, I should've worded it better. I checked x=\frac{\pi}{6} with the second derivative. I got -3 which makes it a local max.

That's correct. However, there are 3 other potential maximums or minimums in the interval to check as well. :)

 

[fghtffyrdmns]

That's correct. However, there are 3 other potential maximums or minimums in the interval to check as well. :)

what! I checked 0, 3pi/4 and pi/2 also. 0 and 3pi/4 I got local maximums for that. But a local minimum at pi/2 :/

 

[fghtffyrdmns]

In the line below, you have divided both sides by cosx, effectively discarding cos x = 0 as a solution to your original equation.

You should have changed the equation above to this:
2sinxcosx - cos x = 0
==> cosx(2sinx - 1) = 0
==> cos x = 0 or sinx = 1/2

Yes! I knew I did that part right. But my teacher just said it was right. By dividing both sides by cos, I would lose a x value - which is a huge mistake. This is how I got pi/2 the first time.

 

[slider142]

what! I checked 0, 3pi/4 and pi/2 also. 0 and 3pi/4 I got local maximums for that. But a local minimum at pi/2 :/

Right. Now the only thing left to do is compare the values of f at those points to find the absolute maximum and the absolute minimum (since your original post has the singular forms of these words, I assume they mean absolute).

 

[fghtffyrdmns]

Whew, you scared me there. I thought you meant that there's more! Yes, these are absolute values (as it is in a restricted domain). The absolute max is at x= pi/6 and absolute min is at x = pi/2.

However, f(0) = 1 and f(pi/2) = 1. But I checked with 2nd derivative and f(pi/2) is the only min. Is this enough proof?

 

[slider142]

Whew, you scared me there. I thought you meant that there's more! Yes, these are absolute values (as it is in a restricted domain). The absolute max is at x= pi/6 and absolute min is at x = pi/2.

However, f(0) = 1 and f(pi/2) = 1. But I checked with 2nd derivative and f(pi/2) is the only min. Is this enough proof?

No. Both 0 and pi/2 are absolute minima, with minimum value 1.
Again, the second derivative only tells you the concavity of a region; it does not tell you whether a value is an absolute minimum or not. It gives only enough information to tell whether a critical point is a local minimum or a local maximum, or possibly neither. It is even possible for the 2nd derivative to fail to give you this information, ie., consider the function f(x) = x⁴ for x = 0. It is obviously a local (and global) minimum, but the second derivative test is inconclusive.

 

[fghtffyrdmns]

No. Both 0 and pi/2 are absolute minima, with minimum value 1.

Hmm. I must've done the 2nd derivative(the calculator part :p) wrong since I got it as a -4 :/

 

[fghtffyrdmns]

No. Both 0 and pi/2 are absolute minima, with minimum value 1.
Again, the second derivative only tells you the concavity of a region; it does not tell you whether a value is an absolute minimum or not. It gives only enough information to tell whether a critical point is a local minimum or a local maximum, or possibly neither. It is even possible for the 2nd derivative to fail to give you this information, ie., consider the function f(x) = x⁴ for x = 0. It is obviously a local (and global) minimum, but the second derivative test is inconclusive.

Damn it. No one told me this :(/ How would I go proving that there is 2 minimas? I know that there is only 1 max.

 

[slider142]

Hmm. I must've done the 2nd derivative(the calculator part :p) wrong since I got it as a -4 :/

The second derivative at x=0 is -4, meaning the graph has negative concavity there (it opens downwards). This does not tell you much about the relation of this point to the values of points in the closed interval under consideration, since this point is not a turning point. Draw a picture.

 

[fghtffyrdmns]

The second derivative at x=0 is -4, meaning the graph has negative concavity there (it opens downwards). This does not tell you much about the relation of this point to the values of points in the closed interval under consideration, since this point is not a turning point. Draw a picture.

You are correct. f(o) = f(pi/2) = 1 but since I know that f"(pi/2) = 2 > 0 it's a min. Logically, f(o) is a min too.