# Showing derivative >0

• April 30th 2012, 07:49 AM
jsndacruz
Showing derivative >0
This problem comes from Edward J. Gaughan's Introduction to Analysis.

The Problem
(4.3.19) Show that $\cos(x) = x^3 + x^2 + 4x$ has precisely one root on the closed interval $[0,\frac{\pi}{2}]$
Attempted Solution
Let $f(x) = x^3 + x^2 + 4x - \cos(x)$. Since each component of $f(x)$ is differentiable and therefore continuous on the interval. Then by the Intermediate Value Theorem, since $f(0) < 0$ and $f(\frac{\pi}{2}) > 0$, $\exists c \in (0, \frac{\pi}{2})$ such that $f(c) = 0$.

Since $\forall x_1,x_2 \in (0,\frac{\pi}{2})$, if $x_1 < x_2$, then $f(x_1), $f(x)$ is monotonic increasing and therefore $\forall x_1,x_2 \in (0,\frac{\pi}{2})$, $g(x) = \frac{f(x_1)-f(x_2)}{x_1-x_2} > 0$.

The issue
The derivative of $f(x)$ at a point $a$ is the limit of $g(x)$ as $x \rightarrow a$. Is there a way to show that this limit is strictly greater than zero? Is it sufficient to find the limit of $g(x)$ at $x=0$ and then say that since $g(x)$ is monotonic increasing, that this would be a lower bound for the derivative of $f(x)$?

EDIT: I should mention that I want to show c is unique. So I'd like to say the limit is strictly greater than zero so that I can apply Rolle's Theorem.
• April 30th 2012, 08:17 AM
Prove It
Re: Showing derivative >0
Quote:

Originally Posted by jsndacruz
This problem comes from Edward J. Gaughan's Introduction to Analysis.

The Problem
(4.3.19) Show that $\cos(x) = x^3 + x^2 + 4x$ has precisely one root on the closed interval $[0,\frac{\pi}{2}]$
Attempted Solution
Let $f(x) = x^3 + x^2 + 4x - \cos(x)$. Since each component of $f(x)$ is differentiable and therefore continuous on the interval. Then by the Intermediate Value Theorem, since $f(0) < 0$ and $f(\frac{\pi}{2}) > 0$, $\exists c \in (0, \frac{\pi}{2})$ such that $f(c) = 0$.

Since $\forall x_1,x_2 \in (0,\frac{\pi}{2})$, if $x_1 < x_2$, then $f(x_1), $f(x)$ is monotonic increasing and therefore $\forall x_1,x_2 \in (0,\frac{\pi}{2})$, $g(x) = \frac{f(x_1)-f(x_2)}{x_1-x_2} > 0$.

The issue
The derivative of $f(x)$ at a point $a$ is the limit of $g(x)$ as $x \rightarrow a$. Is there a way to show that this limit is strictly greater than zero? Is it sufficient to find the limit of $g(x)$ at $x=0$ and then say that since $g(x)$ is monotonic increasing, that this would be a lower bound for the derivative of $f(x)$?

Rolle's Theorem states that in order for a function to pass through the x axis twice, it needs to turn somewhere. Therefore, if there are two roots, \displaystyle \begin{align*} x = a, x = b \end{align*}, then there exists some \displaystyle \begin{align*} c \in (a, b) \end{align*} such that \displaystyle \begin{align*} f'(c) = 0 \end{align*}.

Anyway, your function is \displaystyle \begin{align*} f(x) = x^3 + x^2 + 4x - \cos{x} \end{align*}, its derivative is \displaystyle \begin{align*} f'(x) = 3x^2 + 2x + 4 + \sin{x} \end{align*}.

It's well known that when \displaystyle \begin{align*} x \in \left(0, \frac{\pi}{2}\right), \sin{x} > 0 \end{align*}, and it's also relatively easy to see that \displaystyle \begin{align*} 3x^2 + 2x + 4 > 0 \end{align*} in that region as well.

Therefore, \displaystyle \begin{align*} f'(x) > 0 \end{align*} for all \displaystyle \begin{align*} x \in \left(0, \frac{\pi}{2}\right) \end{align*}, which means the function does not turn, and therefore there can not possibly be any more than one root in that interval.
• April 30th 2012, 08:51 AM
jsndacruz
Re: Showing derivative >0
Thanks Prove It.

The problem is that I can't use the equation for $f'(x)$; I can only use its limit form. I was trying to show the derivative is greater than zero because then I could apply Rolle's Theorem and show that since the derivative was strictly greater than zero, the function could not "turn" to get another zero.
• April 30th 2012, 09:03 AM
Prove It
Re: Showing derivative >0
Quote:

Originally Posted by jsndacruz
Thanks Prove It.

The problem is that I can't use the equation for $f'(x)$; I can only use its limit form. I was trying to show the derivative is greater than zero because then I could apply Rolle's Theorem and show that since the derivative was strictly greater than zero, the function could not "turn" to get another zero.

That's ridiculous. Why would you not be able to use the equation for the derivative? Just some simple analysis of the equation is enough to show that the derivative is positive for all x in that region.
• April 30th 2012, 09:38 AM
jsndacruz
Re: Showing derivative >0
I know, but we're just getting started on formally defining differentiation. It's an introductory analysis course and we're only allowed to use theorems/identities that we can prove. I don't know how to prove that the derivative of $\cos(x)$ is $\sin(x)$ using the limit definition of differentiation, so I can't use that form.
• April 30th 2012, 09:56 AM
Prove It
Re: Showing derivative >0
Quote:

Originally Posted by jsndacruz
I know, but we're just getting started on formally defining differentiation. It's an introductory analysis course and we're only allowed to use theorems/identities that we can prove. I don't know how to prove that the derivative of $\cos(x)$ is $\sin(x)$ using the limit definition of differentiation, so I can't use that form.

Well first of all, the derivative of \displaystyle \begin{align*} \cos{(x)} \end{align*} is NOT \displaystyle \begin{align*} \sin{(x)} \end{align*}, it's \displaystyle \begin{align*} -\sin{(x)} \end{align*}. Anyway...

\displaystyle \begin{align*} f(x) &= \cos{(x)} \\ \\ f(x + h) &= \cos{(x + h)} \\ &= \cos{(x)}\cos{(h)} - \sin{(x)}\sin{(h)} \\ \\ \frac{f(x + h) - f(x)}{h} &= \frac{\cos{(x)}\cos{(h)} - \sin{(x)}\sin{(h)} - \cos{(x)}}{h} \\ &= \frac{\cos{(x)}\left[\cos{(h)} - 1\right] - \sin{(x)}\sin{(h)}}{h} \\ &= \cos{(x)}\left[\frac{\cos{(h)} - 1}{h}\right] - \sin{(x)}\left[\frac{\sin{(h)}}{h}\right] \\ \\ \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} &= \lim_{h \to 0}\left\{ \cos{(x)}\left[\frac{\cos{(h)} - 1}{h}\right] - \sin{(x)}\left[\frac{\sin{(h)}}{h}\right] \right\} \\ &= \cos{(x)}\lim_{h \to 0}\left[\frac{\cos{(h)} - 1}{h}\right] - \sin{(x)}\lim_{h \to 0}\left[\frac{\sin{(h)}}{h}\right] \\ &= \cos{(x)}\lim_{h \to 0}\left\{\frac{\left[\cos{(h)} - 1\right]\left[\cos{(h)} + 1\right]}{h\left[\cos{(h)} + 1\right]}\right\} - \sin{(x)}\cdot 1 \\ &= \cos{(x)}\lim_{h \to 0}\left\{\frac{\cos^2{(h)} - 1}{h\left[\cos{(h)} + 1\right]}\right\} - \sin{(x)} \end{align*}

\displaystyle \begin{align*} &= \cos{(x)}\lim_{h \to 0}\left\{\frac{-\sin^2{(h)}}{h\left[\cos{(h)} + 1\right]}\right\} - \sin{(x)} \\ &= \cos{(x)}\lim_{h \to 0}\left[\frac{\sin{(h)}}{h}\right]\lim_{h \to 0}\left[-\frac{\sin{(h)}}{\cos{(h)} + 1}\right] - \sin{(x)} \\ &= \cos{(x)} \cdot 1 \cdot 0 - \sin{(x)} \\ &= -\sin{(x)} \end{align*}

And if you need proof that \displaystyle \begin{align*} \lim_{h \to 0}\frac{\sin{(h)}}{h} = 1 \end{align*}, see here...
• April 30th 2012, 10:31 AM
jsndacruz
Re: Showing derivative >0
Quote:

Originally Posted by Prove It
Well first of all, the derivative of \displaystyle \begin{align*} \cos{(x)} \end{align*} is NOT \displaystyle \begin{align*} \sin{(x)} \end{align*}, it's \displaystyle \begin{align*} -\sin{(x)} \end{align*}. Anyway..

Twas but a typo.

For the rest of it .. thanks for such a thorough proof. I would have never been able to come up with that myself.