Найдите значение производной функции в точке x₀: 23.21. a) y = x² + 2x - 1, x₀ = 0; 6) y = x³ - 3x + 2, x₀ = -1; в) y = x² + 3x – 4, x₀ = 1; г) y = x³ - 9x² + 7, x₀ = 2. 23.22. a) y = 2/x - 1, x₀ = 4; 6) y = √x + 4, x₀ = 9; в) y = 8/x - 6, x₀ = 1; г) y = √x + 5, x₀ = 4.

Решения:

23.21.

1. a) \( y = x^2 + 2x - 1 \), \( x_0 = 0 \)
   \( y' = 2x + 2 \)
   \( y'(0) = 2(0) + 2 = 2 \)
2. б) \( y = x^3 - 3x + 2 \), \( x_0 = -1 \)
   \( y' = 3x^2 - 3 \)
   \( y'(-1) = 3(-1)^2 - 3 = 3 - 3 = 0 \)
3. в) \( y = x^2 + 3x - 4 \), \( x_0 = 1 \)
   \( y' = 2x + 3 \)
   \( y'(1) = 2(1) + 3 = 5 \)
4. г) \( y = x^3 - 9x^2 + 7 \), \( x_0 = 2 \)
   \( y' = 3x^2 - 18x \)
   \( y'(2) = 3(2)^2 - 18(2) = 12 - 36 = -24 \)

23.22.

1. a) \( y = \frac{2}{x} - 1 \), \( x_0 = 4 \)
   \( y = 2x^{-1} - 1 \)
   \( y' = -2x^{-2} = -\frac{2}{x^2} \)
   \( y'(4) = -\frac{2}{4^2} = -\frac{2}{16} = -\frac{1}{8} \)
2. б) \( y = \sqrt{x} + 4 \), \( x_0 = 9 \)
   \( y = x^{1/2} + 4 \)
   \( y' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \)
   \( y'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{2 \cdot 3} = \frac{1}{6} \)
3. в) \( y = \frac{8}{x} - 6 \), \( x_0 = 1 \)
   \( y = 8x^{-1} - 6 \)
   \( y' = -8x^{-2} = -\frac{8}{x^2} \)
   \( y'(1) = -\frac{8}{1^2} = -8 \)
4. г) \( y = \sqrt{x} + 5 \), \( x_0 = 4 \)
   \( y = x^{1/2} + 5 \)
   \( y' = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \)
   \( y'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{2 \cdot 2} = \frac{1}{4} \)

Ответ: 23.21. а) 2; б) 0; в) 5; г) -24. 23.22. а) -1/8; б) 1/6; в) -8; г) 1/4.