question_answer

Find the slope of the tangent to the curve \[y={{x}^{3}}-3x+2\] at the point whose x coordinate is 3.

Answer:

Given equation of curve \[y={{x}^{3}}-3x+2\]     ?(i) When x = 3, then             \[y={{(3)}^{3}}-3\times 3+2=27-9+2=20\] \[\therefore \] The point on the curve is (3, 20). On differentiating Eq. (i) both sides w.r.t. x, we get             \[\left( \frac{dy}{dx} \right)=3{{x}^{2}}-3\] \[\therefore \] Slope of tangent at (3, 20) \[={{\left( \frac{dy}{dx} \right)}_{(3,\,\,20)}}\]             \[=3{{(3)}^{2}}-3=27-3=24\]