Question

Find the slope of the tangent to the curve y = x^3 - x + 1 at the point where the curve cuts the y-axis A. 1 B-1 C. 2 D. 0​

Answer 1

Answer:

Explanation:

Answer

Correct option is

A

4

B

−4

f(x)=(x+1)(x−3)

Now  

f(x)=0

⇒x=−1,x=3

Differentiating f(x) with respect to x

dx

dy

=x+1+x−3

=2x−2

=2(x−1)

Now slope of the tangent at (h,k) will be  

dx

dy

h,k

Hence slopes of the tangent at x=−1 and x=3, will be

dx

dy

x=−1

=2(x−1)  

x=−1

=−4  

And  

dx

dy

x=3

=2(x−1)  

x=3

​

=4