23.23. Вычислите скорость изменения функции y = g(x) в точке x₀: a) g(x) = x² + 2x, x₀ = 2; 6) g(x) = (√3 + 1)√x, x₀ = 1; в) g(x) = x² + 4√x - 4x, x₀ = 4; г) g(x) = 1/x (4/x - 2), x₀ = -0,5.

Решения:

Скорость изменения функции равна её производной.

23.23.

1. a) \( g(x) = x^2 + 2x \), \( x_0 = 2 \)
   \( g'(x) = 2x + 2 \)
   \( g'(2) = 2(2) + 2 = 6 \)
2. б) \( g(x) = (\sqrt{3} + 1)\sqrt{x} \), \( x_0 = 1 \)
   \( g(x) = (\sqrt{3} + 1)x^{1/2} \)
   \( g'(x) = (\sqrt{3} + 1) \cdot \frac{1}{2}x^{-1/2} = \frac{\sqrt{3} + 1}{2\sqrt{x}} \)
   \( g'(1) = \frac{\sqrt{3} + 1}{2\sqrt{1}} = \frac{\sqrt{3} + 1}{2} \)
3. в) \( g(x) = x^2 + 4\sqrt{x} - 4x \), \( x_0 = 4 \)
   \( g(x) = x^2 + 4x^{1/2} - 4x \)
   \( g'(x) = 2x + 4 \cdot \frac{1}{2}x^{-1/2} - 4 = 2x + \frac{2}{\sqrt{x}} - 4 \)
   \( g'(4) = 2(4) + \frac{2}{\sqrt{4}} - 4 = 8 + \frac{2}{2} - 4 = 8 + 1 - 4 = 5 \)
4. г) \( g(x) = \frac{1}{x} \left(\frac{4}{x} - 2\right) \), \( x_0 = -0.5 \)
   \( g(x) = \frac{4}{x^2} - \frac{2}{x} = 4x^{-2} - 2x^{-1} \)
   \( g'(x) = -8x^{-3} - (-2)x^{-2} = -\frac{8}{x^3} + \frac{2}{x^2} \)
   \( g'(-0.5) = -\frac{8}{(-0.5)^3} + \frac{2}{(-0.5)^2} = -\frac{8}{-0.125} + \frac{2}{0.25} = 64 + 8 = 72 \)

Ответ: а) 6; б) (√3 + 1)/2; в) 5; г) 72.