Concavity, etc for f(x) = x^2 (6 - x^2)^(1/2)

[Guest]

I have worked on this problem again and again. In fact, I have erased all my work and started from scratch again and again. The question is this:

Find all critical points of f(x)=x^2(6-x^2)^(1/2)
Then find any points of inflection and describe the concavity of the function.

I get the first derivative correct without issue. I do not know if I am getting the second derivative correct. My TI-89 gives me a different answer then what I've got. Can anyone help with this?

Thank you!!

 

[pka]

Here is what MathCad gives me.
\(\displaystyle \L
\begin{array}{rcl}
y & = & x^2 \sqrt {6 - x^2 } \\
y' & = & \frac{{ - 3\left( {x^3 - 4x} \right)}}{{\sqrt {6 - x^2 } }} \\
y'' & = & \frac{{ - 6\left( {x^4 - 9x^2 + 12} \right)}}{{\left( {x^2 - 6} \right)\sqrt {6 - x^2 } }} \\
\end{array}\)

 

[stapel]

I get the first derivative correct without issue. I do not know if I am getting the second derivative correct. My TI-89 gives me a different answer then what I've got.

What did you get for the first and second derivatives?

Please provide details. Thank you.

Eliz.

 

[soroban]

Hello, ezrajoelmicah!

My answers agree with pka's MathCad answers.

Find all critical points of: \(\displaystyle \,f(x)\:=\:x^2(6\,-\,x^2)^{\frac{1}{2}}\)

Then find any points of inflection and describe the concavity of the function.

We have: \(\displaystyle \,f(x)\;=\;x^2(6\,-\,x^2)^{\frac{1}{2}}\)

Then: \(\displaystyle \,f'(x)\;=\;x^2\cdot\frac{1}{2}\cdot(6\,-\,x^2)^{-\frac{1}{2}}\cdot(-2x)\,+\,2x\cdot(6\,-\,x^2)^{\frac{1}{2}}\)

Simplify: \(\displaystyle \,f;(x)\;= \;-x^3(6\,-\,x^2)^{-\frac{1}{2}}\,+\,2x(6\,-\,x^2)^{\frac{1}{2}}\)

Factor: \(\displaystyle \,f'(x)\;=\;-(6\,-\,x^2)^{-\frac{1}{2}}\cdot\left[x^3\,-\,2x(6\,-\,x^2)\right]\)

. . . . . . \(\displaystyle f'(x)\;= \;-(6\,-\,x^2)^{-\frac{1}{2}}[3x^3\,-\,12x]\)

Factor: \(\displaystyle \,f'(x)\;=\;-(6\,-\,x^2)^{-\frac{1}{2}}\cdot3\cdot(x^3\,-\,4x)\)

. . . . . . \(\displaystyle f'(x)\;=\;-3(6\,-\,x^2)^{-\frac{1}{2}}(x^3\,-\,4x)\;\;\Rightarrow\;\;\L\frac{-3(x^3\,-\,4x)}{\sqrt{6\,-\,x^2}}\)

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We have: \(\displaystyle \,f'(x)\;= \;-3(x^2\,-\,4x)(6\,-\,x^2)^{-\frac{1}{2}}\)

Then: \(\displaystyle \,f''(x) \;= \;-3\left[(x^3\,-\,4x)\cdot\left(-\frac{1}{2}\right)(6\,-\,x^2)^{-\frac{3}{2}}\cdot(-2x)\,+\,(3x^2\,-\,4)(6\,-\,x^2)^{-\frac{1}{2}}\right]\)

. . . . . . \(\displaystyle f''(x)\;=\;-3\left[x(x^3\,-\,4x)(6\,-\,x^2)^{-\frac{3}{2}} \,+ \,(3x^2\,-\,4)(6\,-\,x^2)^{-\frac{1}{2}}\right]\)

Factor: \(\displaystyle \,f''(x) \;= \;-3(6\,-\,x^2)^{-\frac{3}{2}}\cdot\left[x(x^3\,-\,4x) \,+\,(3x^2\,-\,4)(6\,-\,x^2)\right]\)

Simplify: \(\displaystyle \,f''(x)\;=\;-3(6\,-\,x^2)^{-\frac{3}{2}}\cdot(-2x^4\,+\,18x^2\,-\,24)\)

Factor: \(\displaystyle \,f''(x)\;=\;-3(6\,-\,x^2)^{-\frac{3}{2}}\cdot(-2)(x^4\,-\,9x^2\,+\,12)\)

. . . . \(\displaystyle f''(x)\;=\;6(6\,-\,x^2)^{-\frac{3}{2}}(x^4\,-\,9x^2\,+\,12)\;\;\Rightarrow\;\;\L\frac{6(x^4\,-\,9x^2\,+\,12)}{(6\,-\,x^2)^{\frac{3}{2}}}\)

 

[Guest]

concavity

Thanks All for your help with this!! I was getting the correct 2nd derivative. My first and second derivatives matched the MathCad answers. Your help is greatly appreciated!!!