To find the equation of the tangent to the curve y = 2x - x^3 at the point x = -1, we need to find the slope of the tangent line and the coordinates of the point where it meets the line y = x.
First, let's find the derivative of y with respect to x to obtain the slope of the tangent line:
y = 2x - x^3
dy/dx = 2 - 3x^2
Next, substitute x = -1 into dy/dx to find the slope:
dy/dx = 2 - 3(-1)^2
dy/dx = 2 - 3(1)
dy/dx = 2 - 3
dy/dx = -1
The slope of the tangent line to the curve at x = -1 is -1.
Now, let's find the y-coordinate of the point where the tangent line meets the line y = x. We can substitute the x-coordinate of the point (-1) into the original equation y = 2x - x^3:
y = 2(-1) - (-1)^3
y = -2 - (-1)
y = -2 + 1
y = -1
The y-coordinate of the point where the tangent line meets the line y = x is -1.
Therefore, the tangent line has a slope of -1 and passes through the point (-1, -1). Using the point-slope form of a linear equation, we can write the equation of the tangent line:
y - (-1) = -1(x - (-1))
y + 1 = -1(x + 1)
y + 1 = -x - 1
y = -x - 2
So, the equation of the tangent line is y = -x - 2.
To find the point where the tangent line meets the line y = x, we can set the equations y = -x - 2 and y = x equal to each other:
-x - 2 = x
2x = -2
x = -1
The x-coordinate is -1, and since y = x, the y-coordinate is also -1.
Therefore, the tangent line meets the line y = x at the point (-1, -1).
Find the equation of tangent to the curve y=2x-x^3 at the point x=-1. Where does that tangent meet the line y=x?
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