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

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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!!
 
Here is what MathCad gives me.
\(\displaystyle \L
\begin{array}{rcl}
y & = & x^2 \sqrt {6 - x^2 } \\
y' & = & \frac{{ - 3\left( {x^3 - 4x} \right)}}{{\sqrt {6 - x^2 } }} \\
y'' & = & \frac{{ - 6\left( {x^4 - 9x^2 + 12} \right)}}{{\left( {x^2 - 6} \right)\sqrt {6 - x^2 } }} \\
\end{array}\)
 
ezrajoelmicah said:
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.
Click to expand...
What did you get for the first and second derivatives?

Please provide details. Thank you.

Eliz.
 
Hello, ezrajoelmicah!

My answers agree with pka's MathCad answers.

Find all critical points of:  f(x) = x2(6 − x2)12\displaystyle \,f(x)\:=\:x^2(6\,-\,x^2)^{\frac{1}{2}}

Then find any points of inflection and describe the concavity of the function.
Click to expand...
We have:  f(x)  =  x2(6 − x2)12\displaystyle \,f(x)\;=\;x^2(6\,-\,x^2)^{\frac{1}{2}}

Then:  f′(x)  =  x2⋅12⋅(6 − x2)−12⋅(−2x) + 2x⋅(6 − x2)12\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:  f;(x)  =  −x3(6 − x2)−12 + 2x(6 − x2)12\displaystyle \,f;(x)\;= \;-x^3(6\,-\,x^2)^{-\frac{1}{2}}\,+\,2x(6\,-\,x^2)^{\frac{1}{2}}

Factor:  f′(x)  =  −(6 − x2)−12⋅[x3 − 2x(6 − x2)]\displaystyle \,f'(x)\;=\;-(6\,-\,x^2)^{-\frac{1}{2}}\cdot\left[x^3\,-\,2x(6\,-\,x^2)\right]

. . . . . . f′(x)  =  −(6 − x2)−12[3x3 − 12x]\displaystyle f'(x)\;= \;-(6\,-\,x^2)^{-\frac{1}{2}}[3x^3\,-\,12x]

Factor:  f′(x)  =  −(6 − x2)−12⋅3⋅(x3 − 4x)\displaystyle \,f'(x)\;=\;-(6\,-\,x^2)^{-\frac{1}{2}}\cdot3\cdot(x^3\,-\,4x)

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

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have:  f′(x)  =  −3(x2 − 4x)(6 − x2)−12\displaystyle \,f'(x)\;= \;-3(x^2\,-\,4x)(6\,-\,x^2)^{-\frac{1}{2}}

Then:  f′′(x)  =  −3[(x3 − 4x)⋅(−12)(6 − x2)−32⋅(−2x) + (3x2 − 4)(6 − x2)−12]\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]

. . . . . . f′′(x)  =  −3[x(x3 − 4x)(6 − x2)−32 + (3x2 − 4)(6 − x2)−12]\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:  f′′(x)  =  −3(6 − x2)−32⋅[x(x3 − 4x) + (3x2 − 4)(6 − x2)]\displaystyle \,f''(x) \;= \;-3(6\,-\,x^2)^{-\frac{3}{2}}\cdot\left[x(x^3\,-\,4x) \,+\,(3x^2\,-\,4)(6\,-\,x^2)\right]

Simplify:  f′′(x)  =  −3(6 − x2)−32⋅(−2x4 + 18x2 − 24)\displaystyle \,f''(x)\;=\;-3(6\,-\,x^2)^{-\frac{3}{2}}\cdot(-2x^4\,+\,18x^2\,-\,24)

Factor:  f′′(x)  =  −3(6 − x2)−32⋅(−2)(x4 − 9x2 + 12)\displaystyle \,f''(x)\;=\;-3(6\,-\,x^2)^{-\frac{3}{2}}\cdot(-2)(x^4\,-\,9x^2\,+\,12)

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