Acurve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at a where x=b+1. show that the gradient of the curve at a is 1.

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

Solution:

We need to show that the gradient of the curve at A is 1

Here given that ,

 --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

so, we get

By putting value of point A and putting value of eq 2 we get

Hence proved that the gradient of the curve at A is 1.