Asked by Anonymous
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.
Answers
Answered by
Reiny
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
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
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