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Kaylis [27]
3 years ago
9

F(x)=8x sqrt(x-x^2) Find the exact maximum

Mathematics
1 answer:
Mariana [72]3 years ago
3 0

First take note of the domain of <em>f(x)</em> ; the square root term is defined as long as <em>x</em> - <em>x</em> ² ≥ 0, or 0 ≤ <em>x</em> ≤ 1.

Check the value of <em>f(x)</em> at these endpoints:

<em>f</em> (0) = 0

<em>f</em> (1) = 0

Take the derivative of <em>f(x)</em> :

f(x)=8x\sqrt{x-x^2}=8x\left(x-x^2\right)^{\frac12}

\implies f'(x)=8\left(x-x^2\right)^{\frac12}+4x\left(x-x^2\right)^{-\frac12}(1-2x)=4\left(x-x^2\right)^{-\frac12}\left(2\left(x-x^2)\right)+x(1-2x)\right)=\dfrac{4(3x-4x^2)}{\sqrt{x-x^2}}

For <em>x</em> ≠ 0, we can eliminate the √<em>x</em> term in the denominator:

x\neq0\implies f'(x)=\dfrac{4\sqrt x (3-4x)}{\sqrt{1-x}}

<em>f(x)</em> has critical points where <em>f '(x)</em> is zero or undefined. We know about the undefined case, which occurs at the boundary of the domain of <em>f(x)</em>. Check where <em>f '(x)</em> = 0 :

√<em>x</em> (3 - 4<em>x</em>) = 0

√<em>x</em> = 0   <u>or</u>   3 - 4<em>x</em> = 0

The first case gives <em>x</em> = 0, which we ignore. The second leaves us with <em>x</em> = 3/4, at which point we get a maximum of max{<em>f(x) </em>} = 3√3 / 2.

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