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Lizzie

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

My answer was x=0...is that right?
 
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.
 
Hello, Lizzie!

The absolute minimum value for f(x)=x12(2x)\displaystyle f(x)\:=\:- x^{\frac{1}{2}}(2\,-\,x) on the interval [0,6]\displaystyle [0, 6] occurs at __.

My answer was x=0\displaystyle x = 0 . . . is that right? . . . . no
Did you use the derivative?

We have: .f(x)=2x12+x32\displaystyle f(x)\:=\:-2x^{\frac{1}{2}} + x^{\frac{3}{2}}

Then: .f(x)=x12+32x12=0\displaystyle f'(x)\:=\:-x^{-\frac{1}{2}}\,+\,\frac{3}{2}x^{\frac{1}{2}}\:=\:0

Multiply by x12:    1+32x=0        x=23\displaystyle x^{\frac{1}{2}}:\;\;-1\,+\,\frac{3}{2}x\:=\:0\;\;\Rightarrow\;\;x\,=\,\frac{2}{3}

And: .f(23)=23(223)=63(43)=469\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 f(0)=0\displaystyle f(0)\,=\,0
 
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