# Putnam Exam Challenge (Maximum Value)

1. Aug 7, 2014

Hi,

Let y= |sin(x) + cos(x) + tan(x) + sec(x) + csc(x) + cot(x)|

Find the minimum value of "y" for all real numbers.

Graphing is not allowed, no devices, calculators whatsoever.

Its VERY hard to find where this function = 0 analytically so it is better to take two different approaches.

(1) Assume y > 0 for (-∞, ∞)
(2) Assume y <0 for (-∞, ∞)

Case 1
-----------
y' = cos(x) - sin(x) + sec^2(x) + sec(x)tan(x) - csc(x)cot(x) - csc^2(x)

The obvious step is set y' = 0, that is VERY hard to do as well.

Let u= sin(x)
Let r = sec(x)
Let o = tan(x)
Let h = cot(x)
Let g = csc(x), this y' is the same as

y' = u' - u + r^2 + r' + g' + h'

0 = u' - u + r^2 + r' + g' + h'

u = u' + r^2 + r' + g' + h' [y' = 0 when this takes place]

This is kind of impossible to figure out. So Ill try this,

y' = cos(x) - sin(x) + sec^2(x) + sec(x)tan(x) - csc(x)cot(x) - csc^2(x)

0 = cos(x) - sin(x) + sec^2(x) + sec(x)tan(x) - csc(x)cot(x) - csc^2(x)

0 = cos(x) - sin(x) + sec(x)[sec(x) + tan(x)] - csc(x)[csc(x) + cot(x)]

0 = cos(x) - sin(x) + [sec(x) + tan(x)]/cos(x) - [csc(x) + cot(x)]/sin(x)

0 = cos^2(x) - cos(x)sin(x) + [sec(x) + tan(x)] - cos(x)[csc(x) + cot(x)]/sin(x)

0 = sin(x)cos^2(x) - cos(x) + sin(x)[sec(x) + tan(x)] - cos(x)[csc(x) + cot(x)]

[sec(x) + tan(x)] = [1 + sin(x)]/[cos(x)]

0 = sin(x)cos^2(x) - cos(x) + sin(x)[1 + sin(x)]/[cos(x)] - cos(x)[1+ cos(x)]/[sin(x)]

Thats all I can do for now.

2. Aug 7, 2014

### Staff: Mentor

These problems are good at misdirecting a student and I don't know if this advice will help:

As described the problem starts with sin(x) + cos(x) and so you proceed down that path of complementary functions symmetry.

However, if you notice that there's an inverse symmetry too with sin(x) + csc(x) or tan(x) + cot(x) and cos(x) + sec(x) you might see something there.

3. Aug 7, 2014

### jbunniii

(2) is impossible, since y is the absolute value of some expression. So $y \geq 0$ is guaranteed.
No, here you're assuming that the expression inside the absolute value is always nonnegative. What basis do you have for that assumption? To see why this won't necessarily work, consider a simpler function, $y = |\sin(x)|$. Clearly $y$ takes on a minimum value of zero whenever $x$ is a multiple of $\pi$. But the derivative of $\sin$ is not zero at these points.

One thing you could try, but will probably be extremely messy, is to recognize that $y = |\text{whatever}|$ is minimized at exactly the same points as $y^2 = (\text{whatever})^2$. This will at least allow you to shed the absolute value, but the squaring operation will give you an expression with 36 terms Many of them will probably cancel or simplify, though.

4. Aug 7, 2014

### jbunniii

It might be useful to introduce the function $f(x) = x + 1/x$. Then:

$f(\sin(x)) = \sin(x) + \csc(x)$
$f(\cos(x)) = \cos(x) + \sec(x)$
$f(\tan(x)) = \tan(x) + \cot(x)$
$y = |f(\sin(x)) + f(\cos(x)) + f(\tan(x))|$
$y^2 = \left(f(\sin(x)) + f(\cos(x)) + f(\tan(x))\right)^2 = \ldots$

You can make use of the fact that $f(x)^2 = x^2 + 2 + 1/x^2 = f(x^2) + 2$. Not sure if any of this will help to simplify things.

5. Aug 7, 2014

### slider142

From the beginning, reduce all trigonometric functions to their definitions in terms of sine and cosine only. Adding all terms to create one fraction will allow you to use the Pythagorean identity and the double angle identity for sine to significantly reduce the complexity of the problem.
Ignore the absolute value and take the derivative of the function inside the absolute value bars to find its relative minimums and maximums. If any value is negative, you may take its absolute value to get the corresponding point for y.
The only other points you need to worry about are places where y = 0, which are easily found using the reduced fraction. If, for example, there is any point where y = 0, then 0 is the absolute minimum value of your function, since the absolute value function cannot return any value less than 0. If there are no points where y = 0, you may inspect the critical points as detailed in the previous sentence.

Last edited: Aug 7, 2014
6. Aug 10, 2014

Hi there,

I didnt notice this reply. This is a very good approach.

y = |f(sin(x)) + f(cos(x)) + f(tan(x))|

It is important to see that everything here can be written in terms of sine and cosine.

Ignoring the absolute value

y1 = sin(x) + cos(x) + sin(x)/cos(x) + cos(x)/sin(x) + 1/sin(x) + 1/cos(x)

y1 = [sin^2(x)cos(x) + cos^2(x)sin(x) + sin^2(x) + cos^2(x) + cos(x) + sin(x)]/[sin(x)cos(x)]

sin(x)cos(x) = (1/2)[sin(2x)]

y1 = 2[cos(x) - cos^3(x) + cos^2(x)sin(x) + 1 + cos(x) + sin(x)]/sin(2x)

y1 = 2[sin(x)(1/2)sin(2x) + cos(x)(1/2)sin(2x) + 1 + cos(x) + sin(x)]/[sin(2x)]

y1 = 2[ {(1/2)sin(2x)}[sin(x) + cos(x)] + [sin(x) + cos(x)] + 1]/[sin(2x)]

y1 = [ sin(2x)[sin(x) + cos(x)] + 2[sin(x) + cos(x)] + 2 ]/sin(2x)

Let u = sin(x) + cos(x)

y1 = [ sin(2x) + 2 ] / sin(2x)

The trick now is to somehow adjust sin(2x) in terms of u.

sin(2x) = 2sin(x)cos(x)

sin(x + x) = sin(x)cos(x) + cos(x)sin(x)

Doesnt work =(

EDIT

u = sin(x) + cos(x)

u^2 = [sin(x) + cos(x)][sin(x) + cos(x)] = sin^2(x) + sin(2x) + cos^2

u^2 - 1 = sin(2x)

y1 = [ sin(2x) + 2 ] / sin(2x)

y1 = [ [u^2 - 1] + 2 ]/[u^2 - 1]

y1 = [u^2 - 1]/[u^2 - 1] + 2/[u^2 + 1]

y1 = + 2/[u^2 + 1]

The problem is

y = | + 2/[u^2 + 1]|

Last edited: Aug 10, 2014
7. Aug 10, 2014

Further more, we can try to differentiate FIRST.

y2 = sin(x) + cos(x) + sin(x)/cos(x) + cos(x)/sin(x) + 1/sin(x) + 1/cos(x)

y2' = cos(x) - sin(x) + [cos^2(x) +sin^2(x)]/cos^2(x) + [-sin^2(x) -cos^2(x)]/sin^2(x) - cos(x)/sin^2(x) + sin(x)/cos^2(x)

y2' = cos(x) - sin(x) + 1/cos^2(x) - 1/sin^2(x) - cos(x)/sin^2(x) + sin(x)/cos^2(x)

8. Aug 10, 2014

### disregardthat

Here you get y1 = [ sin(2x) + 2 +2 ] / sin(2x)

From before, you have y1 = [ [u^2 - 1] + 2 +2 ]/[u^2 - 1], and you get

y1 = [u^2 - 1]/[u^2 - 1] + 2/[u^2 - 1] + 2/[u^2-1] = u+[2u+2)]/[u^2-1] = u+2/[u-1].

(not [u^2+1] in the denominator)
--------------------------------------
However, you can not ignore the absolute value sign. We can consider the four different quadrants by themselves $(0,\frac{\pi}{2}),(\frac{\pi}{2},\pi), (\pi,\frac{3\pi}{2}), (\frac{3\pi}{2},2\pi)$. Then by translating by $\frac{\pi}{2}, \pi$ and $\frac{3\pi}{2}$ respectively we can reduce to four different cases where x is only in the interval $(0,\frac{\pi}{2})$ (where sin(x) and cos(x) are positive). Just remember that

$\sin(x+\frac{\pi}{2}) = \cos(x)$
$\cos(x+\frac{\pi}{2}) = -\sin(x)$
$\sin(x+\pi) = -\sin(x)$
$\cos(x+\pi) = -\cos(x)$
$\sin(x+\frac{3\pi}{2}) = -\cos(x)$
$\cos(x+\frac{3\pi}{2}) = \sin(x)$

So, here we have translated the expressions to account for the different intervals:

$|\sin(x)+\cos(x)+\frac{1}{\sin(x)} + \frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}|$ for the quadrant $(0,\frac{\pi}{2})$ (not translating).

$|\cos(x)-\sin(x)+\frac{1}{\cos(x)} - \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} - \frac{\sin(x)}{\cos(x)}|$ for the quadrant $(\frac{\pi}{2},\pi)$ (translating by $\frac{\pi}{2}$).

$|-\sin(x)-\cos(x)-\frac{1}{\sin(x)} - \frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}|$ for the quadrant $(\pi,\frac{3\pi}{2})$ (translating by $\pi$).

$|-\cos(x)+\sin(x)-\frac{1}{\cos(x)} + \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} - \frac{\sin(x)}{\cos(x)}|$ for the quadrant $(\frac{3\pi}{2},2\pi)$ (translating by $\frac{3\pi}{2}$).
----------------------------------------
Now, since $x \in (0,\frac{\pi}{2})$ we know that sin(x) and cos(x) are positive in every one of these expressions, so we may remove the absolute value sign for the quadrant $(0,\frac{\pi}{2})$, and end up with:

$\sin(x)+\cos(x)+\frac{1}{\sin(x)} + \frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}$ for the quadrant $(0,\frac{\pi}{2})$. For quadrants $(\frac{\pi}{2},\pi)$ and $(\frac{3\pi}{2},2\pi)$ we have:

$|\cos(x)-\sin(x)+\frac{1}{\cos(x)} - \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} - \frac{\sin(x)}{\cos(x)}|$ for the quadrant $(\frac{\pi}{2},\pi)$ (translating by $\frac{\pi}{2}$).

and

$|-\cos(x)+\sin(x)-\frac{1}{\cos(x)} + \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} - \frac{\sin(x)}{\cos(x)}|$ for the quadrant $(\frac{3\pi}{2},2\pi)$ (translating by $\frac{3\pi}{2}$).

However, by a substitution $y = \frac{\pi}{2}-x$ we get, since $\sin(y) = \cos(x), \cos(y) = \sin(x)$, that this last expression is just

$|-\cos(y)+\sin(y)-\frac{1}{\cos(y)} + \frac{1}{\sin(y)} - \frac{\cos(y)}{\sin(y)} - \frac{\sin(y)}{\cos(y)}|$ where y is also in $(0,\frac{\pi}{2})$. This is however equivalent to the expression for the quadrant $(\frac{\pi}{2},\pi)$.
-----------------------
It is furthermore obvious that
$|\cos(x)-\sin(x)+\frac{1}{\cos(x)} - \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} - \frac{\sin(x)}{\cos(x)}|$ and
$|-\sin(x)-\cos(x)-\frac{1}{\sin(x)} - \frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}|$
are always strictly less than
$\sin(x)+\cos(x)+\frac{1}{\sin(x)} + \frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}$
(remember that $x \in (0,\frac{\pi}{2})$), and so we know that the first quadrant cannot be where the function is minimized.
--------------------------------
Thus, we reduce to the quadrants $(\frac{\pi}{2},\pi)$ and $(\pi,\frac{3\pi}{2})$ and minimizing the functions

$|\cos(x)-\sin(x)+\frac{1}{\cos(x)} - \frac{1}{\sin(x)} - \frac{\cos(x)}{\sin(x)} - \frac{\sin(x)}{\cos(x)}|$ and
$|-\sin(x)-\cos(x)-\frac{1}{\sin(x)} - \frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)}|$
for $x \in (0,\frac{\pi}{2})$.

For the second function, we see that $\frac{1}{\sin(x)} \geq \frac{\cos(x)}{\sin(x)}$ and $\frac{1}{\cos(x)} \geq \frac{\sin(x)}{\cos(x)}$, so the second function is just

$\sin(x)+\cos(x)+\frac{1}{\sin(x)} + \frac{1}{\cos(x)} - \frac{\sin(x)}{\cos(x)} - \frac{\cos(x)}{\sin(x)}$.

It is not obvious to me that the first function is either always positive or always negative though, even though wolfram alpha suggests it.

9. Aug 10, 2014

### disregardthat

Okay, for the first function, we want to prove that
$-\cos(x)+\sin(x)-\frac{1}{\cos(x)} + \frac{1}{\sin(x)} + \frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{\cos(x)} \geq 0$.
We see that $\frac{\cos(x)}{\sin(x)} \geq \cos(x)$, so we reduce to showing
$\sin(x)+\frac{1}{\sin(x)} + \frac{\sin(x)}{\cos(x)} \geq \frac{1}{\cos(x)}$

Multiplying, this is $\sin^2(x)\cos(x)+\cos(x) +\sin(x)^2 \geq \sin(x)$.
But $\sin^2(x) \cos(x)+\cos(x) \geq 2 \sqrt{\sin^2(x) \cos(x) \cdot \cos(x)} = 2\sin(x)\cos(x)$ (EDIT: we apply the AM-GM inequality), so we need only to show that
$2\sin(x)\cos(x) +\sin(x)^2 \geq \sin(x)$.
Dividing, this is
$2\cos(x)+ \sin(x) \geq 1$. But squaring this, shows that it is immediate.

Hence the second function is just

$f_2(x) = \sin(x)+\frac{1}{\sin(x)}-\cos(x)-\frac{1}{\cos(x)} + \frac{\cos(x)}{\sin(x)} + \frac{\sin(x)}{\cos(x)}$

and the first one is still

$f_1(x) = \sin(x)+\frac{1}{\sin(x)}+\cos(x)+\frac{1}{\cos(x)} - \frac{\cos(x)}{\sin(x)} - \frac{\sin(x)}{\cos(x)}$

The problem remains to minimize these now (for $x \in (0, \frac{\pi}{2})$).

Last edited: Aug 10, 2014
10. Aug 10, 2014

I'm still in high school (not even 11th grade)..... so I have to absorb this and it will take some time, while I do that, can you assist me with this?

There is a specific trig identity that

sin(x) + cos(x) = sqrt(2)*sin(x + pi/4)

This could significantly help for us to minimize the amount of trig functions there.

ONLY STARTING WITH THE LHS

I noticed that sin(x + pi/2) = cos(x)
Also that cos(x + pi/2) = -sin(x)

sin(x) + cos(x) = (-)(-sin(x)) + cos(x) = sin(x + pi/2) - cos(x + pi/2)

But this doesnt help.

I could try sin(x + pi/4) and use the formulas, so

sin(x + pi/4) = sin(x)cos(pi/4) + cos(x)sin(pi/4) = sqrt(2)/2 [sin(x) + cos(x)]

-_- sin(x) + cos(x) = sin(x + pi/4) * sqrt(2)/2

So lets do it the way we did it before.

y = |sin(x) + cos(x) + tan(x) + cot(x) + sec(x) + csc(x)|

Let u = sin(x) + cos(x) But first ignore the absolute sign (temporarily)

y = u + sin(x)/cos(x) + cos(x)/sin(x) + 1/cos(x) + 1/sin(x)

Lets denote s for sine, and c for cosine.

y = s + c + s/c + c/s + 1/c + 1/s

y = [s^2c + c^2s + 1 + s + c]/cs

y = [s^2c + c^2s + u + 1]/cs

The s^2c + c^2s is the main problem.

[s + c]^2 = s^2 + 2cs + c^2
But that doesnt help either. Im just trying several ideas here. We could try cofunctions but those wont help either. Splitting may help here.

y = [s^2c + c^2s + u + 1]/cs
y = u + [u + 1]cs

[s + c]^2 = s^2 + 2cs + c^2

cs = [u^2 - 1]/2

y = u + [u + 1]/cs
y = u + 2[u + 1]/[u^2 - 1]
y = u + 2/[u - 1]

This is the simplest you can get in terms of "u". Still ignoring the absolute value (for now).

y = sin(x) + cos(x) + 2/[sin(x) + cos(x) - 1]

y' = cos(x) - sin(x) + 2* [sin(x) - cos(x)]/[sin(x) + cos(x) -1]^2

Any ideas?

Any ideas for what we could do potentially with u = s + c?

Last edited: Aug 10, 2014
11. Aug 10, 2014

### disregardthat

The final steps are included here, don't look if you want to finish it yourself.
EDIT:removed the spoiler

Using your method (which was a very good idea), we get that while setting $u = \sin(x) + \cos(x)$:
$f_1(x) = u+\frac{2}{u+1}$,
and while setting $v = \sin(x)-\cos(x)$:
$f_2(x) = v+\frac{2}{v+1}$.
So consider the functions $g_1(u) = u+\frac{2}{u+1}, g_2(v) = v+\frac{2}{v+1}$.

We know that $u = \sqrt{2}\sin(x+\frac{\pi}{4})$ and $v = \sqrt{2}\sin(x-\frac{\pi}{4})$ by basic trigonometry. Since $x \in (0,\frac{\pi}{2})$, we see the domain of u is $1 < u \leq \sqrt{2}$, and the domain of v is $-1 < v < 1$ by looking at the unit circle.

Differentiating, we get $g_1^{\prime}(u) = 1-\frac{2}{(u+1)^2}$, which has zeros $\pm \sqrt{2} -1$. But we see that $g_1^{\prime}(u) > 0$ for u in its domain (even including 1), so $g_1(u)$ is strictly increasing, always strictly larger than $g_1(1) = 2$ (on the domain of u).

On the other hand, for $g_2(v)$, we get that the zero $\sqrt{2}-1$ is in the domain of v. It is decreasing for $1 < v < \sqrt{2}-1$, and increasing for $\sqrt{2}-1 < v < 1$. So, when $v = \sqrt{2}-1$, the function $g_2(v)$ is minimized, with value $g_2(\sqrt{2}-1) = 2 \sqrt{2}-1$.

Finally, we easily see that the minimal point $2 \sqrt{2}-1$ of $g_2(v)$ is less than 2, which is what $g_1(u)$ is always larger than. So we conclude that the global minimal value for the trigonometric function in question is $2 \sqrt{2}-1$.

Last edited: Aug 10, 2014
12. Aug 10, 2014

### disregardthat

See my above posts, you were almost there, and would end up with the expression u+2/[u-1].

Upon thinking about it, it was completely unecessary to go through the steps I have . You could of course just work with |u+2/(u-1)|, and consider where it is positive and where it is negative from there. I think you can take it from here, anyway.

13. Aug 10, 2014

Hi,

It is very hard to see because of the spoiler (my internet isnt great).

I noticed something, why might be an errorYou have written

$f_1(x) = u+\frac{2}{u+1}$

But it is supposed to be

$f_1(x) = u+\frac{2}{u-1}$

Where I did the red. Because the $u+1$ on the top cancels with the $u+1$ on the bottom. so you are left with $u-1$ in the fraction.

$f_1(x) = u+\frac{2}{u-1}$

how did you end up with your result there?

Thanks

14. Aug 10, 2014

### disregardthat

I split up my posts, to make it easier to read. Just take it from the first one. I am using the functions f_1 and f_2 here, defined above. I first show that you can ignore the two other cases, reducing to these two functions. I also removed the spoiler tag.

15. Aug 10, 2014

UPDATE

I just read this (after posting previously), and so I did have

f(x) = u + 2/[u-1]

You defined a function in terms of u

g(u) = u + 2/[u-1]

And then differentiated with respect to "u", is this allowed? Since "u" is indirectly a function of "x" the chain rule would state

g'(x) = dg/du * du/dx
du/dx = cos(x) - sin(x)

You cant make it so that the independent variable is "u", since it is a composite function after all that u(x) = cos(x) + sin(x)

Your thoughts? But IF we continue like this,

Let v = cos(x) - sin(x) therefore, du/dx = v

g(x) = u + 2/[u-1]

g'(x) = (1)(v) - (1)(v)/[u-1]^2

0 = v - v/[u-1]^2

0 = v[1 - 1/[u-1]^2]

v = 0
sin(x) = cos(x) ---> $x = pi/4 + 2kpi$

1 - 1/[u-1]^2 = 0

$1 = 1/[u-1]^2$
$(u-1)^2 = 1$
$u -1 = plus-minus 1$
$u = 2 or 0$

Case #1 (u = 2)
$sin(x) + cos(x) = 2$

By definition,
$Rsin(x+t) = Rsin(x)cos(t) + Rsin(t)cos(x)$
It would only make sense if
$Rsin(t) = 1, and Rcos(t) = 1$
$R = plus-minus √2$

We need to find the value of "t" so that we can use the sine addition formula.
$√2sin(t) = 1 & √2cos(t) = 1$
$t = pi/4$
But then there is the negative square root of 2
$t = 5pi/4$

$√2sin(x+pi/4) = sin(x) + cos(x)$

$√2sin(x+pi/4) = 2$
$sin(x + pi/4) = 2/√2$
$x + pi/4 = arcsin(√2)$

This isnt correct so

$sin(x+pi/4) = 0$
$x = -pi/4$

These arent correct, what is wrong?

16. Aug 10, 2014

### Staff: Mentor

Don't forget for these kinds of tests there's the formal way to solve a problem which may be very lengthy and there's an intuitive way to solve the problem which comes from longtime exposure of the math in other problems. The ones who do best on these tests utilize their vast personally acquired knowledge to cut through the morass and ntuit an answer.

THe best example is perhaps Ramanujan who had a great though not perfect intuition and created a number of theorems and conjectures that are still stumping mathematicians today. How he did it is impossible to say but he was able to extropolate a lot of knowledge from a second rate mathematics book then in use in India at the time when he was growing up. In some cases, he even invented his own notation to describe a problem and its solution and it took a mathematician like GH Hardy to see through all that to find his genius.

17. Aug 10, 2014

### disregardthat

Amad, remember that we are trying to find the minimum. We are not required to find the derivative of g_1 in terms of the derivative of f_1. If our function g_1 has the same range (gives the same values) as f_1, then we may restrict our attention to g_1 and forget f_1, since we are only interested in the minimal value.

I'm not sure what you are trying to do when you are differentiating v with x, it is entirely unnecessary. It is also not necessary to relate u and v in any way here. If you have further questions I think you should give a summary of what you've done so far, because it is slightly confusing when you are using the same function names as I have for a different function.

Last edited: Aug 10, 2014
18. Aug 13, 2014

Hi disregardthat,

Let me see the analysis behind what you did.

f(g(x)) where g(x) was u = sin(x) + cos(x) correct?

There is a problem there though,

With g(u) you let u be the INDEPENDENT variable, the graphs of f(x) and g(u) do not coincide.

Give me a second,

You found a critical point in terms of u,

u = √2 - 1 and u = -√2 - 1

This really means,

sin(x) + cos(x) = √2 - 1 & sin(x) + cos(x) = -√2 - 1

You stated

But $g_1$ doesnt have the same range as $f_1$

you found the minimum value for g(u), how's that the minimum value for f(x)?

Thanks

19. Aug 13, 2014

### disregardthat

The graphs may not coincide, but the functions take the same values. Note that we are restricting the domain of g to the values of cos(x) + sin(x). So g is defined for the range of cos(x) + sin(x), thats the only values u may take, and the only place where g is defined. We even discard some of the zeroes of the derivative because of this (and we do not consider them). Yes, we are making u an independent variable afterwards, but only when making sure that g takes the same values as f, by restricting g's domain. Therefore the minimum value of g is the same as the minimum value of f.

20. Aug 13, 2014

That is fascinating, I just graphed both [assuming u was an independent variable] and surprisingly they DO have the same minimum.

It's probably my Algebra knowledge, but why does the range of g depend on sin(x) + cos(x) ?

What property/theorem is this?

EDIT

This doesnt work for every function. Take ln(x^2 + x + 1)

f(x) = ln(x^2 + x + 1)

Let u= x^2 + x

f(u) = ln(u + 1)

They have different ranges.

Last edited: Aug 13, 2014