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.
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
To find the value of the constant b, we need to determine the equation of the derivative of the function j(x) and then solve for b when x = 3.
Step 1: Find the derivative of j(x):
To find the derivative of j(x), we need to use the power rule, the chain rule, and the quotient rule. Let's break down the steps:
j(x) = √(x + 1) + b / x^2
The derivative of j(x) can be calculated as follows:
j'(x) = d/dx(√(x + 1)) + d/dx(b / x^2)
Using the chain rule, d/dx(√(x + 1)) can be written as 1/2 * (x + 1)^(-1/2) * d/dx(x + 1).
Using the power rule, the derivative of b / x^2 will be -2b / x^3.
Applying the derivative rules, we have:
j'(x) = 1/2 * (x + 1)^(-1/2) * d/dx(x + 1) - 2b / x^3
The derivative of x + 1 with respect to x is simply 1.
Therefore, the equation of the derivative j'(x) becomes:
j'(x) = 1/2 * (x + 1)^(-1/2) - 2b / x^3
Step 2: Evaluate the derivative at x = 3 to find the value of b:
Now that we have the equation of the derivative j'(x), we can substitute x = 3 into this equation:
j'(3) = 1/2 * (3 + 1)^(-1/2) - 2b / 3^3
Simplifying further:
j'(3) = 1/2 * 4^(-1/2) - 2b / 27
= 1/2 * (1/2) - 2b / 27
= 1/4 - 2b / 27
Since the function j(x) has a critical point at x = 3, the derivative j'(3) must equal zero:
j'(3) = 0
1/4 - 2b / 27 = 0
To solve for b, we can multiply both sides of the equation by 27:
27/4 - 2b = 0
Now, isolate b by subtracting 27/4 from both sides:
-2b = -27/4
Finally, divide both sides by -2:
b = (-27/4) / -2
b = 27/8
Therefore, the value of the constant b is 27/8.