Differentiation

Question

A curve has the equation y= (x-a)√(x-b) for x≥b, where a and b are constants. It intersects the x-axis at the point A with x=b+1. Show that the gradient of the curve of A is 1.

 

 

Thank You!

Answer 1

The x-intercept from the equation y= (x-a)√(x-b), x = a or x = b

From the x= b+1, where b is constant, therefore only x = a is true.

Now, dy/dx = √(x-b) + (x-a)/(2√(x-b)), where x - b = 1, and  x - a = 0

        dy/dx = 1

Therefore the gradient of the curve is 1