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[Lizzie]

The absolute minimum value for f(x) = - x^½(2 - x) on the interval [0, 6] occurs at: ?

My answer was x=0...is that right?

 

[stapel]

Sorry, but I can't read the exponent on the first "x". Is that a minus one-half?

When you reply, please include at least an outline of the steps you did. Thank you.

Eliz.

 

[soroban]

Hello, Lizzie!

The absolute minimum value for \(\displaystyle f(x)\:=\:- x^{\frac{1}{2}}(2\,-\,x)\) on the interval \(\displaystyle [0, 6]\) occurs at __.

My answer was \(\displaystyle x = 0\) . . . is that right? . . . . no

Did you use the derivative?

We have: .\(\displaystyle f(x)\:=\:-2x^{\frac{1}{2}} + x^{\frac{3}{2}}\)

Then: .\(\displaystyle f'(x)\:=\:-x^{-\frac{1}{2}}\,+\,\frac{3}{2}x^{\frac{1}{2}}\:=\:0\)

Multiply by \(\displaystyle x^{\frac{1}{2}}:\;\;-1\,+\,\frac{3}{2}x\:=\:0\;\;\Rightarrow\;\;x\,=\,\frac{2}{3}\)

And: .\(\displaystyle f\left(\frac{2}{3}\right)\:=\:-\sqrt{\frac{2}{3}}\,\left(2\,-\,\frac{2}{3}\right)\:=\:-\frac{\sqrt{6}}{3}\left(\frac{4}{3}\right)\:=\:-\frac{4\sqrt{6}}{9}\)

. . This is lower than your \(\displaystyle f(0)\,=\,0\)