Question

Найдите уравнение касательной к графику
1) y=x²-6x+5, x₀ =2
2) y = x₂+8x-9, x₀ = -2
3) y = eˣ, x₀ = 2
4) y = 2ˣ, x₀ =1

Answer 1

Ответ:

1) f(2) = 4 - 24 + 5 = -15

f'(x) = 2x - 6

f'(2) = 4 - 6 = -2

y = -15 - 2(x - 2) = - 15 - 2x + 4 = -11 - 2x

2) f(-2) = 1/4 - 16 - 9 = 1/4 - 25 = -24,75

f'(x) = 2x + 8

f'(-2) = -4 + 8 = 4

y = -24,75 + 4(x + 2) = -24,75 + 4x + 8 = -16,75 + 4x

3) f(2) = e^2

f'(x) = e^x

f'(2) = e^2

y = e^2 + e^2 * (x - 2) = e^2 + e^2 * x - 2e^2 = -e^2 + e^2 * x

4) f(1) = 2^1= 2

f'(x) = 2^x * ln2

f'(1) = 2 * ln2

y = 2 * ln2 + 2(x - 1) = 2 * ln2 + 2x - 2