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Sidorovich
@Sidorovich
July 2022
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Найти уравнение общей касательной к графику функций f(x)=x^2+4x+8 и g(x)=x^2+8x+4
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ShirokovP
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Пусть прямая y = kx + b - искомая общая касательная к графикам функций. x1 и x2 - абсциссы соответствующих точек касания
y = f(x0) + f'(x0) (x - x0) = f'(x0) * x + f(x0) - f ' (x0)*x0
f ' (x0) = tga = k
k = p ' (x1) = g ' (x2)
p ' (x) = 2x + 4, p '(x1) = 2x1 + 4
g ' (x) = 2x + 8, g'(x2) = 2x2 + 8
2x1 + 4 = 2x2 + 8
x1 + 2 = x2 + 4
b = p(x1) - p ' (x1)*x1 = x1^2 + 4x1 + 8 - (2x1 + 4)*x1 =
= x1^2 + 4x1 + 8 - 2x1^2 - 4x1 = - x1^2 + 8
b = g(x2) - g'(x2) * x2 = x2^2 + 8x2 + 4 - (2x2 + 8)*x2 =
= x2^2 + 8x2 + 4 - 2x2^2 - 8x2 = - x2^2 + 4
- x1^2 + 8 = - x2^2 + 4
Решим систему
x1 + 2 = x2 + 4
- x1^2 + 8 = - x2^2 + 4
x1 - x2 = 2
x1^2 - x^2 = 4
x1 - x2 = 2
(x1 - x2)(x1 + x2) = 4
x1 - x2 = 2
2*(x1 + x2) = 4
x1 - x2 = 2
x1 + x2 = 2
------------------- +
2x1 = 4
x1 = 2
x2 = 2 - x1 = 2 - 2 = 0
k = p '(x1) = 2x1 + 4 = 2*2 + 4 = 4 + 4 = 8
b = - x1^2 + 8 = - 2^2 + 8 = 8 - 4 = 4
Получаем
y = 8x + 4
2 votes
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Sidorovich
Что такое p и BMW
Sidorovich
Кхм... p и b
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Answers & Comments
Verified answer
Пусть прямая y = kx + b - искомая общая касательная к графикам функций. x1 и x2 - абсциссы соответствующих точек касанияy = f(x0) + f'(x0) (x - x0) = f'(x0) * x + f(x0) - f ' (x0)*x0
f ' (x0) = tga = k
k = p ' (x1) = g ' (x2)
p ' (x) = 2x + 4, p '(x1) = 2x1 + 4
g ' (x) = 2x + 8, g'(x2) = 2x2 + 8
2x1 + 4 = 2x2 + 8
x1 + 2 = x2 + 4
b = p(x1) - p ' (x1)*x1 = x1^2 + 4x1 + 8 - (2x1 + 4)*x1 =
= x1^2 + 4x1 + 8 - 2x1^2 - 4x1 = - x1^2 + 8
b = g(x2) - g'(x2) * x2 = x2^2 + 8x2 + 4 - (2x2 + 8)*x2 =
= x2^2 + 8x2 + 4 - 2x2^2 - 8x2 = - x2^2 + 4
- x1^2 + 8 = - x2^2 + 4
Решим систему
x1 + 2 = x2 + 4
- x1^2 + 8 = - x2^2 + 4
x1 - x2 = 2
x1^2 - x^2 = 4
x1 - x2 = 2
(x1 - x2)(x1 + x2) = 4
x1 - x2 = 2
2*(x1 + x2) = 4
x1 - x2 = 2
x1 + x2 = 2
------------------- +
2x1 = 4
x1 = 2
x2 = 2 - x1 = 2 - 2 = 0
k = p '(x1) = 2x1 + 4 = 2*2 + 4 = 4 + 4 = 8
b = - x1^2 + 8 = - 2^2 + 8 = 8 - 4 = 4
Получаем
y = 8x + 4