absolute min/max values of y = x^2 + 2x^2/3, y = 1/(x^2 + 2x

[endoo]

Find the absolute min and max values:

1) y = x^2 + 2x^2/3 on the interval [-2, 2]

2) y = 1/(x^2 + 2x + 5) on the interval [-2, 1]

 

[stapel]

So you've taken the derivatives, set them equal to zero, found the critical points, tested the interval endpoints, and... then what? Where are you stuck?

Please be specific. Thank you.

Eliz.

 

[endoo]

well i know the derivative of the first is:

y=2x + 4/3x^1/3 but now i dont know how to solve for x

then for the second: im not even sure how to do the derivative for that one

 

[soroban]

Re: absolute min/max values of y = x^2 + 2x^2/3, y = 1/(x^2

Hello, endoo!

I understand your difficulty . . .
Even if you handled the derivative in #1, the rest is quite tricky.

Find the absolute min and max values:

$\displaystyle 1)\;y \:= \:x^2\,+\,2x^{\frac{2}{3}}$ on the interval $\displaystyle [-2, 2]$

The derivative equation is: $\displaystyle \:y'\:=\:2x\,+\,\frac{4}{3}x^{-\frac{1}{3}}\;=\;0$

We have: $\displaystyle \:2x\;=\;-\frac{4}{3}x^{-\frac{1}{3}}$

Multiply by $\displaystyle x^{\frac{1}{3}}: \;\;2x^{\frac{4}{3}} \;=\;-\frac{4}{3}\;\;\Rightarrow\;\;x^{\frac{4}{3}}\;=\;-\frac{2}{3}$

Raise both sides to the power $\displaystyle \frac{3}{4}:$
. . $\displaystyle \left(x^{\frac{4}{3}}\right)^{\frac{3}{4}} \;= \;\left(-\frac{2}{3}\right)^{\frac{3}{4}}\;\;\Rightarrow\;\;x\:=\:\left(-\frac{2}{3}\right)^{\frac{3}{4}}$

But $\displaystyle \,\left(-\frac{2}{3}\right)^{\frac{3}{4}}\:=\:\left[\left(-\frac{2}{3}\right)^3\right]^{\frac{1}{4}} \:=\:\sqrt[4]{-\frac{8}{27}}$ . . . not a real number!
Hence, there are no horizontal tangents on that interval.


I tested the endpoints.
. . \(\displaystyle \text{At }x\,=\,-2:\;y\:=\-2)^2\,+\,2(-2)^{\frac{2}{3}} \:=\:4 + 2\sqrt[3]{4}\)
. . \(\displaystyle \text{At }x\,=\,2:\;y\:=\2)^2\,+\,2(-2)^{\frac{2}{3}} \:=\:4 + 2\sqrt[3]{4}\)

The endpoints are at the same "height", but the graph is not a horizontal line.
There must be an extreme point somewhere on the interval . . . but where?

Then I saw it . . . The derivative is: \(\displaystyle \L\,y'\:=\:2x\,+\,\frac{4}{3\sqrt[3]{x}}\)
When $\displaystyle x\,=\,0$, the slope is undefined . . . There is a vertical tangent.

When $\displaystyle x\,=\,0,\;y\:=\:0^2 + 2(0)^{\frac{2}{3}} \:=\:0$ . . . The origin (0,0) is on the graph.

There is a "cusp" at the origin.
The curve might be shaped like this:

                  |
        *         |         *
              *   |   *
                * | *
                 *|*
                  |
      - + - - - - * - - - - + -
       -2         |         2

Therefore: $\displaystyle \:\begin{array}{cc}\text{Abs.max } & \left(\pm2,\:4+2\sqrt[3]{4}\right) \\ \text{Abs.min } & (0,0)\end{array}$

\(\displaystyle 2)\;y \:= \:\L\frac{1}{x^2\,+\,2x\,+\,5}\) on the interval $\displaystyle [-2, 1]$

We have: \(\displaystyle \,y\:=\x^2\,+\,2x\,+\,5)^{-1}\)

Then: $\displaystyle \,y' \:=\:-(x^2\,+\,2x\,+\,5)^{-2}(2x\,+\,2)\:=\:-\frac{2(x\,+\,1)}{(x^2\,+\,2x\,+\,5)^2}$

And $\displaystyle y'\,=\,0$ when $\displaystyle x\,=\,-1$

The second derivative (after a lot of simplifying) is: \(\displaystyle \L\,y''\;=\;\frac{2(x^2\,+\,6x\,-\,1)}{(x^2\,+\,2x\,+\,5)^3}\)

At $\displaystyle x\,=\,-1:\;y''\,=\,-\frac{3}{16}$ . . . negative: concave down $\displaystyle \cap$
. . Hence, there is a relative maximum at $\displaystyle \left(-1,\,\frac{1}{4}\right)$

Test endpoints:
. . At $\displaystyle x\,=\,-2:\;y\,=\,\frac{1}{5}$
. . At $\displaystyle x\,=\,1:\;y\,=\,\frac{1}{8}$

Therefore: $\displaystyle \:\begin{array}{cc}\text{Abs.max } & \left(-1,\,\frac{1}{4}\right) \\ \text{Abs.min } & \left(1,\,\frac{1}{8}\right) \end{array}$