At which point is the tangent to the curve y | vedclass

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(D) Let the point be $(x_1, y_1)$.Given the curve $y = x^3 + 5$.The slope of the tangent is $\frac{dy}{dx} = 3x^2$.At point $(x_1, y_1)$,the slope is

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$(1, 6), (-1, 4)$

Vedclass · Answer

(D) Let the point be $(x_1, y_1)$.Given the curve $y = x^3 + 5$.The slope of the tangent is $\frac{dy}{dx} = 3x^2$.At point $(x_1, y_1)$,the slope is $m_1 = 3x_1^2$.The given line is $x + 3y = 2$,which can be written as $3y = -x + 2$ or $y = -\frac{1}{3}x + \frac{2}{3}$.The slope of this line is $m_2 = -\frac{1}{3}$.Since the tangent is perpendicular to the line,$m_1 	imes m_2 = -1$.Therefore,$(3x_1^2) 	imes (-\frac{1}{3}) = -1$.$-x_1^2 = -1 \implies x_1^2 = 1 \implies x_1 = \pm 1$.If $x_1 = 1$,then $y_1 = (1)^3 + 5 = 6$.If $x_1 = -1$,then $y_1 = (-1)^3 + 5 = 4$.Thus,the points are $(1, 6)$ and $(-1, 4)$.

At which point is the tangent to the curve $y = x^3 + 5$ perpendicular to the line $x + 3y = 2$?

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