Decreasing the volume increases the pressure, which shifts the equilibrium to the side with fewer moles of gas, favoring the production of NH3.

Increasing the pressure will shift the equilibrium to the side with fewer moles of gas, which is the right side (N2O4) in this case.

Adding more I2 increases its concentration, which shifts the equilibrium to the right to produce more HI.

Adding more HI increases the concentration of products, which shifts the equilibrium to the left to form more reactants.

For an exothermic reaction, increasing the temperature shifts the equilibrium to the left, favoring the reactants.

For an exothermic reaction, increasing the temperature shifts the equilibrium to the left, favoring the reactants, as the system tries to absorb the added heat.

According to Le Chatelier's principle, increasing the temperature of an exothermic reaction shifts the equilibrium to the left, favoring the reactants.

The vertex x-coordinate is given by -b/(2a) = -4/(2*-1) = 2.

The vertex x-coordinate is found using x = -b/(2a) = -6/(-2) = 3.

The minimum occurs at x = 2. f(2) = 2(2^2) - 8(2) + 10 = 2.

Set f'(x) = 0. f'(x) = 6x^2 - 18x + 12 = 0. Critical points are x = 2.

f''(x) = 12x - 18. Setting f''(x) = 0 gives x = 1.5. The inflection point is (1.5, f(1.5)).

f'(x) = 6x^2 - 18x + 12. The critical points are x = 1 and x = 2. Test intervals show increase in (0, 3).

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 3. Testing intervals shows f is increasing on (1, 3).

f'(x) = 6x^2 - 18x + 12. Setting f'(x) = 0 gives x = 1 and x = 2. f(1) = 5 is a local maximum.

Find f'(x) = 6x^2 - 18x + 12, set to 0. Local maxima at x = 2 gives f(2) = 6.

The vertex is at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 7 = -1. So, the vertex is (2, -1).

The vertex occurs at x = 2. f(2) = 3(2^2) - 12(2) + 7 = -5.

The x-coordinate of the vertex is given by x = -b/(2a) = 12/(2*3) = 2.

The vertex occurs at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 9 = 3.

The vertex x-coordinate is found using -b/(2a) = 12/(2*3) = 2.

f''(x) = 18x - 24. Setting f''(x) = 0 gives x = 4/3. This is the inflection point.

Finding f'(x) = 9x^2 - 24x + 9 and solving gives two critical points. The second derivative test confirms one maximum and one minimum.

To find the point of inflection, we compute f''(x) = e^x - 2. Setting f''(x) = 0 gives e^x = 2, leading to x = ln(2). The closest integer is x = 1.

f(x) = ln(x) is not defined for x ≤ 0, hence not differentiable at x = 0.

f'(x) = cos(x) - sin(x). Setting f'(x) = 0 gives tan(x) = 1, so x = π/4 + nπ. In [0, 2π], the maximum occurs at x = 3π/4.

f'(x) = cos(x) - sin(x). At x = π/4, f'(π/4) = cos(π/4) - sin(π/4) = √2/2 - √2/2 = 0.

f'(x) = 2x + 2.

The function is a polynomial and is differentiable everywhere.

f'(x) = 2x + 2. Setting f'(x) = 0 gives x = -1. f(-1) = 1.