Question

Find the slope of the tangent to curve y = x3 − x + 1 at the point whose x-coordinate is 2

Answer 1

The given curve is y=x3-x+1.

=$\frac{dy}{dx}$=3x2-1

The slope of the tangent to a curve at $(x_0, y_0)$ is $(\frac{dy}{dx})\bigg] _{(x_0,y_0)}$.

It is given that x0 = 2.

Hence, the slope of the tangent at the point where the x-coordinate is 2 is given by,

$(\frac{dy}{dx}) \bigg]_{x=2}$=3x2-1]x=2=3(2)2-1=12-1=11.