To find the acute angle that the curve y = 1-3x² cuts the x-axis, we need to find the slope of the tangent line at the point where the curve intersects the x-axis.
When the curve intersects the x-axis, y = 0. Therefore, we need to solve the equation 0 = 1 - 3x² for x:
3x² = 1 x² = 1/3 x = ±sqrt(1/3)
We can see that there are two points where the curve intersects the x-axis: (-sqrt(1/3), 0) and (sqrt(1/3), 0).
To find the slope of the tangent line at each of these points, we need to take the derivative of y with respect to x:
dy/dx = -6x
At (-sqrt(1/3), 0), the slope of the tangent line is dy/dx = -6(-sqrt(1/3)) = 2sqrt(3).
At (sqrt(1/3), 0), the slope of the tangent line is dy/dx = -6(sqrt(1/3)) = -2sqrt(3).
The acute angle between the x-axis and a line with slope m is given by arctan(m). Therefore, the acute angle that the tangent lines at these points make with the x-axis are:
arctan(2sqrt(3)) ≈ 75° and arctan(-2sqrt(3)) ≈ -75°
Since we're looking for the acute angle, the answer is 75°, which is option C.