If <em>x</em>² + <em>y</em>² = 1, then <em>y</em> = ±√(1 - <em>x</em>²).
Let <em>f(x)</em> = |<em>x</em>| + |±√(1 - <em>x</em>²)| = |<em>x</em>| + √(1 - <em>x</em>²).
If <em>x</em> < 0, we have |<em>x</em>| = -<em>x</em> ; otherwise, if <em>x</em> ≥ 0, then |<em>x</em>| = <em>x</em>.
• Case 1: suppose <em>x</em> < 0. Then
<em>f(x)</em> = -<em>x</em> + √(1 - <em>x</em>²)
<em>f'(x)</em> = -1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = -1/√2 → <em>y</em> = ±1/√2
• Case 2: suppose <em>x</em> ≥ 0. Then
<em>f(x)</em> = <em>x</em> + √(1 - <em>x</em>²)
<em>f'(x)</em> = 1 - <em>x</em>/√(1 - <em>x</em>²) = 0 → <em>x</em> = 1/√2 → <em>y</em> = ±1/√2
In either case, |<em>x</em>| = |<em>y</em>| = 1/√2, so the maximum value of their sum is 2/√2 = √2.
In order to reduce ANY fraction to lowest terms, find any common factors
of the numerator and denominator, and divide them both by it. If they still
have a common factor, then divide them by it again. Eventually, they won't
have any common factor except ' 1 ', and then you'll know that the fraction is
in lowest terms.
Do 15 and 40 have any common factors ?
Let's see . . .
The factors of 15 are 1, 3, <em>5</em>, and 15 .
The factors of 40 are 1, 2, 4,<em> 5</em>, 8, 10, 20, and 40 .
Ah hah ! Do you see that ' <em>5</em> ' on both lists ? That's a common factor.
So 15/40 is NOT in lowest terms.
Divide the numerator and denominator both by 5 :
15 / 40 =<em> 3 / 8</em>
3 and 8 don't have any common factor except ' 1 '.
So 3/8 is the same number as 15/40, but in lowest terms.
Answer:
1. The second number line, the third number line, the first number line.
2.
