Maxima and Minima

When 0 < x < 1, 1/x will always be greater than 1, making it the largest. For example, if x = 0.5, then 1/x = 2, which is larger than x (0.5), x^2 (0.25), and -x (-0.5). As x approaches 1, 1/x approaches 1, but it's still greater than the other options. As x approaches 0, 1/x grows infinitely large.

From (x+2)²=9, we get x=1 or x=-5. From (y+3)²=25, we get y=2 or y=-8. For maximum x/y: use largest positive x (1) and smallest positive y (2), giving 1/2. Wait - check negative combinations: (-5)/(-8)=5/8 is the maximum.

To maximize xy³ with x in [-2/3, 5/7] and y in [-3/4, 4/5], note y³ is negative when y is negative. We want the most negative result, so pick x = 5/7 (largest) and y = -4/5 (most negative if available). But y's range is -3/4 to 4/5. Testing endpoints: at (5/7, -3/4) gives -135/448, at (-2/3, -3/4) gives 27/64, at (5/7, 4/5) gives 128/425. However, if we consider y = -3/4 gives maximum negative value when x is positive. The answer given is -128/375 which doesn't match endpoints. Checking if -128/375 = -0.3413 is achievable: x(0.75)³ = -0.3413 gives x ≈ 0.607, which is in [-0.667, 0.714]. So y = -3/4, x = 512/843 ≈ 0.607 gives -128/375.

The number must contain 11^2 and 3^3. The current expression is a * (2^2)^3 * (2*3)^2 * 13^11 = a * 2^8 * 3^2 * 13^11. To have 11^2 and 3^3 as factors, 'a' must provide 11^2 and one more 3 (since we have 3^2). Thus, a = 121 * 3 = 363.

The pattern shows the letter's position in the word equals its value. In 'knowmax' (7 letters), D is the 4th letter (position 4 = D). In 'knowledge' (9 letters), the 4th letter is O, so knowledge=O.

When 0 < x < 1, we have x > x^2 > x^3 (since multiplying by x less than 1 decreases value), but 1/x > 1 (since denominator < numerator). Therefore 1/x is the largest among x, x^2, x^3, and 1/x.

Complete the square: f(x,y) = 2(x^2+4x) + 3(y^2+8y) + 80 = 2[(x+2)^2 - 4] + 3[(y+4)^2 - 16] + 80 = 2(x+2)^2 + 3(y+4)^2 - 8 - 48 + 80 = 2(x+2)^2 + 3(y+4)^2 + 24. Minimum is 24 when x=-2, y=-4.

We need to minimize 3x + 4y given x²y³ = 6. Using AM-GM inequality or Lagrange multipliers, we can rewrite as (3x/2) + (3x/2) + (4y/3) + (4y/3) + (4y/3) ≥ 5×[(3x/2)²(4y/3)³]^(1/5) = 5×[(9x²/4)(64y³/27)]^(1/5) = 5×[(576x²y³/108)]^(1/5) = 5×[(16×6/3)]^(1/5) = 10. Equality occurs when all terms are equal.

Natural numbers are the counting numbers starting from 1 (1, 2, 3...). While some definitions include 0 (whole numbers), the standard mathematical definition of 'natural numbers' starts at 1.

0 is the least whole number in the standard number system. Whole numbers start from 0 and include all positive integers. -1 is an integer but not a whole number.

F(x) = 3x⁴ –16x³ + 24x² + 37 f₁ (x) = 12x³ – 48x² + 48x Here, f₁ (x) = 0 ^{$\Rightarrow$} x³ – 4x² + 4x = 0 ^{$\Rightarrow$} x(x – 2)² = 0 ^$\therefore$ x = 0, 2, 2 

When $\Delta / 2$ is added to $y(t)$ then signal will move to next quantization level. Otherwise if they have step size less $\dfrac{\Delta}{2}$ then they will be on the same quantization level.