Question

If y= ax-b / (x-1) (x-4) has a turning point at A(2,-1), find the values of a and b so that y has maxima at the point A.

Answer 1

Answer:

a = 1 , b = 0

Step-by-step explanation:

Given the curve y has a turning point at A(2, -1). This implies

1) Curve passes through A(2, -1) i.e., y(2) = -1

2) As we know, turning point is the point on the curve at which slope of the curve changes from positive to negative or negative to positive

 If slope changes from positive to negative, it is point of maxima,

else if slope changes from negative to positive it is point of minima.

At turning point, we have slope = 0, dy/dx = 0

Given y= ax-b / (x-1) (x-4)

We need to differentiate and equate dy/dx to 0, for simplification purpose lets take logarithm on both sides,  

Taking log on both sides we get

 log y = log(ax-b / (x-1) (x-4)) = log(ax-b)-log(x-1)-log(x-4)

 Differentiating above equation w.r.t x, we get

(dy/dx)/y = a/ax-b - 1/x-1 - 1/x-4--------(*),

but dy/dx =0 at A(2, -1)

a/2a-b -1 +1/2 = 0

On solving we get, b = 0.

Since curve passes through A(2, -1) , point A should satisfy the given equation. On substitution we get

-1 = 2a/(2-1)(2-4)

=> a = 1.