# Math Help - Show that the function is monotonic

1. ## Show that the function is monotonic

Show that the function is monotonic on the interval $]0,\frac{\pi}{4}[$.

$sinx + cosx + tanx + cotx$

I created another topic in Trigonometry about solving the derivative for this function, and it seems the derivative is a bit complicated. Are there any other ways to solve this that doesn't require finding the roots of the derivative (I can't find all the roots)?

2. What does monotonic mean? It means that the function is never decreasing or never increasing. Therefore, a monotonic function must satisfy $f'(x)\le 0$ OR $f'(x)\ge 0$

3. Originally Posted by colby2152
What does monotonic mean? It means that the function is never decreasing or never increasing. Therefore, it never has a solution to $f'(x)=0$
Is that true of $f(x) = x^3$?

4. Originally Posted by Plato
Is that true of $f(x) = x^3$?
Good point Plato. I wasn't totally comfortable with my statement, and you find the flaw. I should have said that values of the derivative must be greater than OR equal to zero.

The following properties are true for a monotonic function f : R → R:

* f has limits from the right and from the left at every point of its domain;
* f has a limit at infinity (either ∞ or −∞) of either a real number, ∞, or −∞.
* f can only have jump discontinuities;
* f can only have countably many discontinuities in its domain.

These properties are the reason why monotonic functions are useful in technical work in analysis. Two facts about these functions are:

* if f is a monotonic function defined on an interval I, then f is differentiable almost everywhere on I, i.e. the set of numbers x in I such that f is not differentiable in x has Lebesgue measure zero.
* if f is a monotonic function defined on an interval [a, b], then f is Riemann integrable.

An important application of monotonic functions is in probability theory. If X is a random variable, its cumulative distribution function

FX(x) = Prob(X ≤ x)

is a monotonically increasing function.

A function is unimodal if it is monotonically increasing up to some point (the mode) and then monotonically decreasing.

5. I know what a monotonic function is. I just need to prove that this particular function is monotonic, and I don't know how to do that.

6. Originally Posted by Spec
I know what a monotonic function is. I just need to prove that this particular function is monotonic, and I don't know how to do that.
$\frac{d}{dx} ( sinx + cosx + tanx + cotx ) = cos(x) - sin(x) + sec^2(x) +- csc^2(x)$

Is this always positive or negative on your interval? (I'd say the easiest way to show this is to graph the thing.)

-Dan

7. We're not supposed to use any tools like calculators, so graphing it is not an acceptable solution.

$cosx$ is monotonically decreasing on the interval.
$-sinx$ is monotonically decreasing on the interval.
$\frac{1}{cos^2x}$ is monotonically increasing on the interval.
$-\frac{1}{sin^2x}$ is monotonically increasing on the interval.

I don't really know what to do with that.

8. Originally Posted by Spec
We're not supposed to use any tools like calculators, so graphing it is not an acceptable solution.

$cosx$ is monotonically decreasing on the interval.
$-sinx$ is monotonically decreasing on the interval.
$\frac{1}{cos^2x}$ is monotonically increasing on the interval.
$-\frac{1}{sin^2x}$ is monotonically increasing on the interval.

I don't really know what to do with that.
You don't need a calculator to graph a function. Calculate the values on your own and plot them. I would plot $f(0), f(\frac{\pi}{6}), f(\frac{\pi}{4})$.

9. As Dan was showing you:

$\frac{d}{dx} ( sinx + cosx + tanx + cotx ) = cos(x) - sin(x) + sec^2(x) +- csc^2(x)$

Now, try to solve this algebraically:

$cos(x) - sin(x) + sec^2(x) - csc^2(x) \le 0$

OR

$cos(x) - sin(x) + sec^2(x) - csc^2(x) \ge 0$

------------------------

$cos(x) - sin(x) + sec^2(x) - csc^2(x) \le 0$

$cos(x) - sin(x) + \frac{1}{cos^2(x)} - \frac{1}{sin^2(x)} \le 0$

$cos^3(x)sin^2(x)-cos^2(x)sin^3(x)+sin^2(x)-cos^2(x) \le 0$

$sin^2(x)(cos^3(x)+1) \le cos^2(x)(sin^3(x)+1)$, looks true to me!

OR

$sin^2(x)(cos^3(x)+1) \ge cos^2(x)(sin^3(x)+1)$, FALSE: Try x = 0

In that interval $[0, \frac{\pi}{4}]$ that you provided, both the cosine and sine function are positive. Therefore, both of those inequalities are true. Sine minus cosine is never more than one when the signs are the same. The function is monotonically decreasing.

10. Originally Posted by colby2152
$sin^2(x)(cos^3(x)+1) \le cos^2(x)(sin^3(x)+1)$, looks true to me!
I don't really know if "looks true to me" constitues a proof.

I still consider this to be unsolved. Solving this inequality is the hard part:

$cosx - sinx + \frac{sin^2x - cos^2x}{cos^2x \cdot sin^2x} < 0, \forall x \in \left]0, \frac{\pi}{4}\right[$

11. $
\begin{gathered}
f'(x) = \cos x - \sin x + \frac{{\sin ^2 x - \cos ^2 x}}
{{sin^2 x\cos ^2 x}} \hfill \\
\hfill \\
\leftrightarrow \left( {\cos x - \sin x} \right)\left( {1 - \frac{{\sin x + \cos x}}
{{sin^2 x\cos ^2 x}}} \right) \hfill \\
\end{gathered}$

now lets look at the following function:

$g(x) = 1 - \frac{{\sin x + \cos x}}
{{sin^2 x\cos ^2 x}}$

notice that:

$\begin{gathered}
\mathop {\lim }\limits_{x \to 0} g(x) = - \infty \hfill \\
g\left( {\frac{\pi }
{4}} \right) < 0 \hfill \\
\end{gathered}$

you can easily verify that g(x) is continues at the half open interval (0,pi/4]
thus if we show that g(x) has no extrema points (maximum) in this interval, it must be negative throughout this interval.

$

g'(x) = - \frac{{\left( {\cos x - \sin x} \right)sin^2 x\cos ^2 x - 2\left( {\sin x\cos ^3 x - \cos x\sin ^3 x} \right)\left( {\sin x + \cos x} \right)}}
{{(...)^2 }} =

$

$
= - \frac{{\left( {\cos x - \sin x} \right)sin^2 x\cos ^2 x - 2\sin x\cos x\left( {\cos ^2 x - \sin ^2 x} \right)\left( {\sin x + \cos x} \right)}}
{{(...)^2 }} =
$

$\begin{gathered}
= - \frac{{\left( {\cos x - \sin x} \right)sin^2 x\cos ^2 x - 2\sin x\cos x\left( {\cos x - \sin x} \right)\left( {\sin x + \cos x} \right)^2 }}
{{(...)^2 }} = \hfill \\
\hfill \\
= - \frac{{\left( {\cos x - \sin x} \right)\sin x\cos x\left[ {\sin x\cos x - 2\left( {\sin x + \cos x} \right)^2 } \right]}}
{{(...)^2 }} = \hfill \\
\end{gathered}$

$
\begin{gathered}
= - \frac{{\left( {\cos x - \sin x} \right)\sin x\cos x\left( { - 3\sin x\cos x - 2} \right)}}
{{(...)^2 }} = \hfill \\
\hfill \\
= \frac{{\left( {\cos x - \sin x} \right)\sin x\cos x\left( {3\sin x\cos x + 2} \right)}}
{{(...)^2 }} \hfill \\
\end{gathered}

$

now let's look at the constituting parts of this expression:

1. $
\left( {\cos x - \sin x} \right)

$

is positive since cosx is bigger than sinx for all x in the given interval.

2. $
{\sin x\cos x}

$

both sinx and cos x are positive in this interval.

3.
${\left( {3\sin x\cos x + 2} \right)}$

trivial...
4. the denominator is squared therefore it's also positive.

therefore g(x) is strictly negative in this interval and therefore f'(x) is strictly negative in this interval since:

$
f'(x) = g(x)\left( {\cos x - \sin x} \right)

$

12. Originally Posted by Spec
Solving this inequality is the hard part:

$\cos x - \sin x + \frac{\sin^2x - \cos^2x}{\cos^2x \cdot \sin^2x} < 0, \forall x \in \left]0, \frac{\pi}{4}\right[$
Notice that this expression is zero when x = π/4. Write it as $f'(x) = \cos x - \sin x + \frac1{\cos^2x} - \frac1{\sin^2x}$, and differentiate it once more, to get

. . . . . $f''(x) = -\sin x - \cos x +\frac{2\sin x}{\cos^3x} + \frac{2\cos x}{\sin^3x} = \frac{(2-\cos^3x)\sin^4x + (2-\sin^3x)\cos^4x}{\sin^3x\cdot\cos^3x}$.

This is clearly positive, so f'(x) is increasing in the interval (0,π/4). But since it is zero at x=π/4, it must be negative when 0<x<π/4.

So f'(x) < 0 and therefore f is monotonically decreasing in that interval.

13. I haven't read all answers, but you don't have to find the roots of f'(x). You just have to show that f'(x) never crosses the x-axis at that specific interval. For example, if you can't show directly that f'(x) is all negative or all positive in that interval, you could for example look at f'(0). If you see that f'(0) is negative, and you somehow can show that f''(0) is all negative in the interval, then f'(0) will only decrease and hence not cross the x-axis in that interval. Hence f(x) will be monotonic.