Finding Absolute Maxima and Minima

[hank]

Ok, my problem is that I'm supposed to find the absolute max and absolute min of the following problem:


\(\displaystyle 1 + \left| {9 - x^2 } \right|\) on the interval [-5,1]

Now, I think what's screwing me up is how to take the derivative of an absolute value. I suspect that's the problem.

I just took the derivative for
f(x) = 1 + 9 - x^2 and f(x) = 1 - 9 + x^2. Maybe you're supposed to apply the chain rule somehow to absolute values?


Anyway, I ended up with this:

f(x) = 2x or -2x

Both of which result in x = 0.

Since it's absolute value, to evaluate you need to plug in for x for
f(x) = 1 + 9 - x^2 and f(x) = 1 - 9 + x^2, right?

When I do that, I get the wrong answers though for the minima.
I get the correct answer for the max: 17 @ x = -5.

Min is supposed to be: 1 @ x = -3.

 

[pka]

\(\displaystyle \L
\frac{{d\left| {f(x)} \right|}}{{dx}} = \frac{ (f(x)){f'(x)}}{{\left| {f(x)} \right|}}\)

Graph the function and you will see.

 

[soroban]

Hello, hank!

I made a sketch and baby-talked my way through it . . .

Find the absolute max and absolute min of: \(\displaystyle \,f(x)\:=\:1\,+\,|9\,-\,x^2|\)
on the interval \(\displaystyle [-5,1]\)

I get the correct answer for the max: \(\displaystyle 17\) at \(\displaystyle x\,=\,-5\)

Minimum is supposed to be: \(\displaystyle 1\) at \(\displaystyle x\,=\,-3\)

The graph of \(\displaystyle f(x)\:=\:9\,-\,x^2\) is a down-opening parabola.
Its vertex is \(\displaystyle (0,9)\); its \(\displaystyle x\)-intercepts are \(\displaystyle (\pm3,\,0).\)

The graph of \(\displaystyle f(x)\:=\:|9\,-\,x^2|\) with everything below the \(\displaystyle x\)-axis
. . reflected upward.

                      |
          *          ***        *
                  *   |   *
                *     |     *
             * *      |      * *
                      |
        ------*-------+-------*------
             -3       |       3

The graph of \(\displaystyle f(x)\:=\:|9\,-\,x^2|\,+\,1\) is moved up one unit.
And we are limited to the inverval \(\displaystyle [-5,1]\)

                      |
          *          ***
          :       *   |   *
          :     *     |   :
          : *  *      |   :
          :           |   : 
          :   *       |   :
          :   :       |   :
      ----+---+-------+---+----
         -5  -3       | - 1

We can see that the absolute minimum is: \(\displaystyle \,(-3,1)\)


The absolute maximum is at \(\displaystyle x\,=\,0\) or \(\displaystyle x\,=\,-5.\)

Since \(\displaystyle f(0)\,=\,10\) and \(\displaystyle f(-5)\,=\,17\)

. . the absolute maximum is \(\displaystyle \,(-5,17).\)

 

[hank]

\(\displaystyle \L
\frac{{d\left| {f(x)} \right|}}{{dx}} = \frac{{f'(x)}}{{\left| {f(x)} \right|}}\)

Graph the function and you will see.

Thanks tons, but how did you get that answer? I've never had to find the derivative of an absolute value before.

 

[pka]

how did you get that answer? I've never had to find the derivative of an absolute value before.

Recall that \(\displaystyle \left| {f(x)} \right| = \sqrt {f^2 (x)}.\)
Find the derivative of the right side.

 

[hank]

Ok, got it. Thanks!