The points at which the tangent passes through the origin for the curve y=4x3−2x5 are

The equation of the tangent line to the curve at a point (x, y) is given by:

y - y1 = m(x - x1)

where (x1, y1) is a point on the curve and m is the slope of the tangent line. The slope of the tangent line is given by the derivative of the function at that point.

The derivative of y = 4x^3 - 2x^5 is:

dy/dx = 12x^2 - 10x^4

So the equation of the tangent line is:

y - (4x^3 - 2x^5) = (12x^2 - 10x^4)(x - x1)

We want to find the points where this line passes through the origin. The origin is the point (0, 0), so we set x = 0 and y = 0 in the equation of the tangent line and solve for x1:

0 - (40^3 - 20^5) = (120^2 - 100^4)(0 - x1)

Solving this equation gives x1 = 0.

Substituting x1 = 0 into the equation of the curve y = 4x^3 - 2x^5 gives y1 = 0.

So the points at which the tangent passes through the origin for the curve y = 4x^3 - 2x^5 are (0, 0).