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## Inequalities – Graphical and Algebraic

I’m teaching a Calculus course this year and came across an interesting problem the other day.  We were working from the Briggs & Cochran Calculus book and discussing the quotient rule.  Problem #40 on page 127 says given the function:

$f(t)=\frac {3t^2}{t^2+1}$

a) Find the values of $t$ for which the slope of the curve is zero.

b) Does the graph of $f(t)$ have a slope of $3$ at any point?

OK – so we need to apply the quotient rule and find $f'(t)$

$f'(t)= \frac{(t^2+1)(6t)-(3t^2)(2t)}{(t^2+1)^2}$

OR

$f'(t)= \frac{6t^3+6t-6t^3}{(t^2+1)^2}$

$f'(t)= \frac{6t}{(t^2+1)^2}$

OK – so part (a) is easy enough.  The derivative is equal to zero when $x=0$.  But what about part (b) ?  We can graph the derivative and see that it’s always less than $3$, but what about algebraically?

The problem can be posed in a number of different ways.  I chose to set it up as:

$3> \frac{6t}{(t^2+1)^2}$

so

$3(t^2+1)^2 > 6t$

and

$(t^2+1)^2 > 2t$

Again, we can graph this and SEE clearly that the graph of $(t^2+1)^2$ is definitely above the graph of $2t$, but I like to know what’s happening algebraically.  In fact, I used to give my College Algebra students algebraic inequalities as an introductory project, so I really wanted to figure this out.

Of course, this came up in the middle of class, so we looked at the graph and I began to consider the algebraic reasoning, but didn’t get far, so I told the class that I would think about it while they had their quiz – but still to no avail.  I worked on it during my free time yesterday and made some progress, then when I came in this morning it all fell together:

$t=0$

$\frac{6t}{(t^2+1)^2}=0$

OK, done there.

$t<0$

$6t<0$

$(t^2+1)^2>0$

$\frac{6t}{(t^2+1)^2}<0$

Again – ok, done.  What was the sticking point for me yesterday were the positive values of $t$. So:

$t>0$

$(t-1)^2\geq0$

$t^2-2t+1\geq0$

So, $t^2+1\geq2t$

Also, $t^2+1>1$

So, $(t^2+1)(t^2+1)>1(t^2+1)\geq2t$

So now we know that $(t^2+1)^2>2t$ and can work our way back to

$3> \frac{6t}{(t^2+1)^2}$

For some reason, I find this beautiful!