+2446 \((-1)^{\frac{1}{2}}\Rightarrow\sqrt{-1}\) is a law of fractional exponents. In general terms, \(x^{\frac{a}{b}}=\sqrt[b]{x^a}=\left(\sqrt[b]{x}\right)^a\)
Why is this the case? Well, I can attempt to explain it to you.
The law of exponents explains how to handle multiplication of exponents with identical bases.
\(x^3*x^2=x^{3+2}=x^5\)
Let's try another example but with fractional exponents.
\(3^{\frac{1}{2}}*3^{\frac{1}{2}}=3^{\frac{1}{2}+\frac{1}{2}}=3^1=3\)
In this example here, \(3^{\frac{1}{2}}\) is a number that when multiplied by itself yields 3. That sounds like the definition of the square root, doesn't it? Therefore, \(3^{\frac{1}{2}}=\sqrt{3}\). One can also expound upon this.
| \(3^{\frac{1}{3}}*3^{\frac{1}{3}}*3^{\frac{1}{3}}=3^{{\frac{1}{3}}+{\frac{1}{3}}+{\frac{1}{3}}}=3^1=3\) |
| \(3^{\frac{1}{4}}*3^{\frac{1}{4}}*3^{\frac{1}{4}}*3^{\frac{1}{4}}=3^{\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}}=3^1=3\) |
| \(3^{\frac{1}{5}}*3^{\frac{1}{5}}*3^{\frac{1}{5}}*3^{\frac{1}{5}}*3^{\frac{1}{5}}=3^{\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}}=3^1=3\) |
| \(3^{\frac{1}{x}}*3^{\frac{1}{x}}*3^{\frac{1}{x}}*...*3^{\frac{1}{x}}*3^{\frac{1}{x}}=3^{\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+...+\frac{1}{x}+\frac{1}{x}}=3^1=3\) |
The pattern continues. \(3^{\frac{1}{3}}\) is a number that when multiplied by itself 3 times yields 3. This is the definition of the cubic root. Then, as you can see, I made a generalization. \(3^{\frac{1}{x}}\), when multiplied x times, yields 3.
+9481 (x + 3)2 + (y - 3)2 = 6 Let's solve this for y .
(y - 3)2 = 6 - (x + 3)2
y - 3 = ±√[ 6 - (x + 3)2 ]
y = ±√[ 6 - (x + 3)2 ] + 3 So....using this value for y....
\(\frac{y}{x}\,=\,\frac{\pm\sqrt{6-(x+3)^2}+3}{x}\)
We can say
\(Y\,=\,\frac{\pm\sqrt{6-(x+3)^2}+3}{x}\) and we want to know the maximum Y value here.
We can get an approximation by looking at a graph. (about 5.828)
If you take the derivative and set it = 0, you will get x = √2 - 2
Then plug this in for x and we can find that the exact maximum value = 3 + 2√2
