Question

Find a point on the curve $y=x^3-3x$ where the tangent is parallel to the chord joining $(1,-2)$ and $(2,2)$.

Answer 1

$y=x^3-3x........\left(1\right)$

$m_1=\left(\frac{dy}{dx}\right)=3x^2-3$

$m_2$ by point $(1,-2),(2,2)$

$m_2=\frac{2+2}{2-1}=4$

parallel

$\begin{array}{c}3x^2-3=4\\ x=\pm \sqrt{\frac{7}{3}}\end{array}$

Put in eq$\left(1\right)$

$y={\left(\sqrt{\frac{7}{3}}\right)}^3-3\left(\sqrt{\frac{7}{3}}\right)$

$y={\left(-\sqrt{\frac{7}{3}}\right)}^3+3\left(\sqrt{\frac{7}{3}}\right)$