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

Solution:

For a curve y = f(x) containing the point (x₁,y₁) the equation of the tangent line to the curve at (x₁,y₁) is given by 

y − y₁ = f′(x₁) (x − x₁)

The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line.

The given curve is

y = x³ - x + 1

Therefore,

dy/dx = d/dx x³ - x + 1

= 3x² - 1

Now,

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

dy/dx]_{x = 2} = 3x² - 1]_{x = 2}

= 3(2)² - 1

= 12 - 1

= 11

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NCERT Solutions Class 12 Maths - Chapter 6 Exercise 6.3 Question 3

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

Summary:

The slope of the tangent to curve y = x³ - x + 1 at the point whose x-coordinate is 2 is 11. The slope of a line is nothing but the change in y coordinate with respect to the change in x coordinate of that line