1. Составьте уравнение касательной к графику функции f(x) в точке графика с хо a) f(x) = x⁴; xo = 2 б) f(x)= -2x; x = -1 в) f(x) = √x² + 3x; x = 1

a) $$f(x) = x^4; x_0 = 2$$

1. $$f(x_0) = f(2) = 2^4 = 16$$
2. $$f'(x) = 4x^3$$
3. $$f'(x_0) = f'(2) = 4 \cdot 2^3 = 4 \cdot 8 = 32$$
4. $$y = 16 + 32(x - 2) = 16 + 32x - 64 = 32x - 48$$

Ответ: $$y = 32x - 48$$

б) $$f(x) = \frac{x^2}{2} - 2x; x_0 = -1$$

1. $$f(x_0) = f(-1) = \frac{(-1)^2}{2} - 2 \cdot (-1) = \frac{1}{2} + 2 = 2.5$$
2. $$f'(x) = x - 2$$
3. $$f'(x_0) = f'(-1) = -1 - 2 = -3$$
4. $$y = 2.5 - 3(x - (-1)) = 2.5 - 3(x + 1) = 2.5 - 3x - 3 = -3x - 0.5$$

Ответ: $$y = -3x - 0.5$$

в) $$f(x) = \sqrt{x^2 + 3x}; x_0 = 1$$

1. $$f(x_0) = f(1) = \sqrt{1^2 + 3 \cdot 1} = \sqrt{1 + 3} = \sqrt{4} = 2$$
2. $$f'(x) = \frac{1}{2\sqrt{x^2 + 3x}} \cdot (2x + 3) = \frac{2x + 3}{2\sqrt{x^2 + 3x}}$$
3. $$f'(x_0) = f'(1) = \frac{2 \cdot 1 + 3}{2\sqrt{1^2 + 3 \cdot 1}} = \frac{5}{2 \cdot 2} = \frac{5}{4} = 1.25$$
4. $$y = 2 + 1.25(x - 1) = 2 + 1.25x - 1.25 = 1.25x + 0.75$$

Ответ: $$y = 1.25x + 0.75$$