absolute min/max values of y = x^2 + 2x^2/3, y = 1/(x^2 + 2x
- Thread starter endoo
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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.
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:
Therefore: \(\displaystyle \:\begin{array}{cc}\text{Abs.max } & \left(\pm2,\:4+2\sqrt[3]{4}\right) \\ \text{Abs.min } & (0,0)\end{array}\)
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: \(\displaystyle \:\begin{array}{cc}\text{Abs.max } & \left(-1,\,\frac{1}{4}\right) \\ \text{Abs.min } & \left(1,\,\frac{1}{8}\right) \end{array}\)
Hello, endoo!
I understand your difficulty . . .
Even if you handled the derivative in #1, the rest is quite tricky.
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:
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-2 | 2Therefore: \(\displaystyle \:\begin{array}{cc}\text{Abs.max } & \left(\pm2,\:4+2\sqrt[3]{4}\right) \\ \text{Abs.min } & (0,0)\end{array}\)
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: \(\displaystyle \:\begin{array}{cc}\text{Abs.max } & \left(-1,\,\frac{1}{4}\right) \\ \text{Abs.min } & \left(1,\,\frac{1}{8}\right) \end{array}\)