Question

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

Answer 1

given curve , y = x³ - x + 1
we have to find slope of tangent of curve at the point whose x - co-ordinate is 2. e.g,. x = 2

we know, slope of tangent of curve y = f(x) at a point (a,b) is 1st order derivatives at that given point. e.g., $\textbf{slope of tangent}=\frac{dy}{dx}|_{(a,b)}$

y = x³ - x + 1
differentiate with respect to x ,
dy/dx = 3x² - 1
at x = 2 , dy/dx = 3(2)² - 1 = 12 - 1 = 11
hence, slope of tangent of curve at x = 2 is 11