Question

^^graph of the equation y=x^3 is shown in the fig then slop of the curve at x=1m will be

Answer 1

1

-4

Explanation:

We have: y=x

2

−

x

2

1

=x

2

−x

−2

Now differentiate,

dx

dy

=2x+2x

−3

So slope of tangent at (−1,0) is

dx

dy

∣

∣

∣

∣

∣

x=−1

=2(−1)+2(−1)

3

=−2−2=−4

Hence slope of normal at the same point is =

4

1