F(x)=3-4/sin^2x M(-Π/4;3Π/4)

Предыдущий
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вопрос

To find the equation of the tangent line to the curve of the function F(x) at point M(-π/4, 3π/4), we first need to find the derivative of the function F(x).

Given F(x) = 3 - 4/sin^2(x), we can rewrite it as F(x) = 3 - 4csc^2(x), where csc(x) represents the cosecant function.

Now, we can find the derivative F'(x) using the chain rule:

F'(x) = d/dx [3 - 4csc^2(x)]
F'(x) = 0 - 4 (d/dx[csc^2(x)])
F'(x) = -4 (-2csc(x) * cot(x))
F'(x) = 8csc(x)cot(x)

Therefore, the derivative of F(x) is F'(x) = 8csc(x)cot(x).

Now, to find the slope of the tangent line at point M(-π/4, 3π/4), we substitute x = -π/4 into the derivative F'(x):

F'(-π/4) = 8csc(-π/4)cot(-π/4)
F'(-π/4) = 8/sec(-π/4) * -1
F'(-π/4) = -8

Hence, the slope of the tangent line at point M is -8.

To find the equation of the tangent line, we can use the point-slope form of a linear equation:

y - y1 = m(x - x1)

Plugging in the values of the point M(-π/4, 3π/4) and the slope m = -8:

y - 3π/4 = -8(x + π/4)

Now simplify to get the equation in the slope-intercept form:

y = -8x - 5π/4

Therefore, the equation of the tangent line to the curve of the function F(x) at point M(-π/4, 3π/4) is y = -8x - 5π/4.