Use both!
You want to minimize <em>P</em>, so differentiate <em>P</em> with respect to <em>x</em> and set the derivative equal to 0 and solve for any critical points.
<em>P</em> = 8/<em>x</em> + 2<em>x</em>
d<em>P</em>/d<em>x</em> = -8/<em>x</em>² + 2 = 0
8/<em>x</em>² = 2
<em>x</em>² = 8/2 = 4
<em>x</em> = ± √4 = ± 2
You can then use the second derivative to determine the concavity of <em>P</em>, and its sign at a given critical point decides whether it is a minimum or a maximum.
We have
d²<em>P</em>/d<em>x</em>² = 16/<em>x</em>³
When <em>x</em> = -2, the second derivative is negative, which means there's a relative maximum here.
When <em>x</em> = 2, the second derivative is positive, which means there's a relative minimum here.
So, <em>P</em> has a relative maximum value of 8/(-2) + 2(-2) = -8 when <em>x</em> = -2.
Answer:
Step-by-step explanation:
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Answer:
<u>start at 5 on the number line and move 8 to the left.</u>
Step-by-step explanation:
Answer:
<h2>Cubing both sides of an equation is reversible.</h2>
Step-by-step explanation:
Squaring both sides of an equation is irreversible, because the square power of negative number gives a positive result, but you can't have a negative base with a positive number, given that the square root of a negative number doesn't exist for real numbers.
In case of cubic powers, this action is reversible, because the cubic root of a negative number is also a negative number. For example
![\sqrt[3]{x} =-1](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%20%3D-1)
We cube both sides
![(\sqrt[3]{x} )^{3} =(-1)^{3} \\x=-1](https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7Bx%7D%20%29%5E%7B3%7D%20%3D%28-1%29%5E%7B3%7D%20%5C%5Cx%3D-1)
If we want to reverse the equation to the beginning, we can do it, using a cubic root on each side
![\sqrt[3]{x}=\sqrt[3]{-1} \\\sqrt[3]{x}=-1](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%3D%5Csqrt%5B3%5D%7B-1%7D%20%5C%5C%5Csqrt%5B3%5D%7Bx%7D%3D-1)
There you have it, cubing both sides of an equation is reversible.
