Find the point on the curve y=x^{3}-11x+5 at | vedclass

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(A) The equation of the given curve is $y=x^{3}-11x+5$. The equation of the tangent is given as $y=x-11$,which is in the form $y=mx+c$. Thus,the slope

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$(2,-9)$ and $(-2,19)$

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(A) The equation of the given curve is $y=x^{3}-11x+5$. The equation of the tangent is given as $y=x-11$,which is in the form $y=mx+c$. Thus,the slope of the tangent is $m=1$. The slope of the tangent to the curve at any point $(x, y)$ is given by the derivative $\frac{dy}{dx} = 3x^{2}-11$. Equating the slope of the tangent to the derivative,we get: $3x^{2}-11 = 1$ $3x^{2} = 12$ $x^{2} = 4$ $x = \pm 2$. When $x=2$,$y = (2)^{3}-11(2)+5 = 8-22+5 = -9$. When $x=-2$,$y = (-2)^{3}-11(-2)+5 = -8+22+5 = 19$. Therefore,the required points are $(2,-9)$ and $(-2,19)$.

Find the point on the curve $y=x^{3}-11x+5$ at which the tangent is $y=x-11$.

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