The function j(x)=sqrt(x+1)+b/x^2 has a critical (turning) point at x = 3. Determine the value for the constant b. Show your work.

Answer 1

j(x)=√(x+1)+b/x^2 = (x+1)^(1/2) + bx^-2

j ' (x) = (1/2)(x+1)^(-1/2) - 2bx^-3 = 1/(2√(x+1)) - 2b/x^3
= 0 when x = 3

(1/2)/√4 - 2b/27 = 0

solve for b