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  • Find the slope of the tangent to curve y = x ^3 – x + 1 at the point whose x-coordinate is 2.

3) Find the slope of the tangent to curvey = x ^3 - x +1 at the point whose x-coordinate is 2.

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Given curve is,
y = x ^3 - x +1
The slope of the tangent at x = 2 is given by
\left ( \frac{dy}{dx} \right )_{x=2} = 3x^2 - 1 = 3(2)^2 - 1= 3\times 4 - 1 = 12 - 1 = 11
Hence, the slope of the tangent at point x = 2 is 11

Posted by

Gautam harsolia

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