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.