For the curve y=4x^{3}-2x^{5},find all the po | vedclass

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(A) The equation of the curve is $y=4x^{3}-2x^{5}$.The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = 12x^{2}-10x^{4}$.The eq

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$(0,0), (1,2), (-1,-2)$

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(A) The equation of the curve is $y=4x^{3}-2x^{5}$.The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = 12x^{2}-10x^{4}$.The equation of the tangent at $(x, y)$ is $Y-y = (12x^{2}-10x^{4})(X-x)$.Since the tangent passes through the origin $(0,0)$,we substitute $X=0$ and $Y=0$ into the equation:$-y = (12x^{2}-10x^{4})(-x)$$y = 12x^{3}-10x^{5}$.Equating this to the original curve equation $y=4x^{3}-2x^{5}$:$12x^{3}-10x^{5} = 4x^{3}-2x^{5}$$8x^{3}-8x^{5} = 0$$8x^{3}(1-x^{2}) = 0$.This gives $x=0$ or $x^{2}=1$,so $x=0, 1, -1$.For $x=0$,$y=4(0)^{3}-2(0)^{5} = 0$.For $x=1$,$y=4(1)^{3}-2(1)^{5} = 2$.For $x=-1$,$y=4(-1)^{3}-2(-1)^{5} = -2$.Thus,the points are $(0,0), (1,2),$ and $(-1,-2)$.

For the curve $y=4x^{3}-2x^{5}$,find all the points at which the tangent passes through the origin.

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