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# Find the critical numbers of the function.$F(x) = x^\frac{4}{5} (x - 4)^2$

## \begin{aligned}F(x) &=x^{4 / 5}(x-4)^{2} \Rightarrow \\F^{\prime}(x) &=x^{4 / 5} \cdot 2(x-4)+(x-4)^{2} \cdot \frac{4}{5} x^{-1 / 5}=\frac{1}{5} x^{-1 / 3}(x-4)[5 \cdot x \cdot 2+(x-4) \cdot 4] \\&=\frac{(x-4)(14 x-16)}{5 x^{1 / 5}}=\frac{2(x-4)(7 x-8)}{5 x^{1 / 5}}\end{aligned}

Derivatives

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### Video Transcript

Okay. We know that we can use the product. Roll off one GPAs FT one to find the derivative were fracturing out the two, as you can see. Okay, Now we have X minus form parentheses. Their first set, X Honest four equals to zero on the end, up with one of our critical numbers. X equals four. Next, we know that we can set seven x months. Eight equal to zero and we end up with X equals eight over seven as our second critical number. And then lastly, don't forget that we have zero as our third critical number because of the denominator.