Find the critical numbers of y=x(4-x)^1/2.
The answer is x=8/3 but I got x=7/2.
y = x√(4-x)
y' = 1√(4-x) + x * 1/2 * 1/√(4-x) * -1
= √(4-x) - x/(2√(4-x))
= (2(4-x) - x)/(2√(4-x))
= (8-3x)/(2√(4-x))
assuming x≠4, we just have to find where the numerator is zero.
x = 8/3
To find the critical numbers of the function y = x(4-x)^(1/2), we need to find the values of x at which the derivative of the function is equal to zero or undefined.
First, let's find the derivative of the function with respect to x. We can use the product rule for differentiation:
y = x(4-x)^(1/2)
Using the product rule, we have:
dy/dx = (d/dx)(x) * (4-x)^(1/2) + x * (d/dx)((4-x)^(1/2))
The derivative of x with respect to x is simply 1, and the derivative of (4-x)^(1/2) with respect to x can be found using the chain rule. Applying the chain rule, we get:
dy/dx = (4-x)^(-1/2) + x * (1/2)(4-x)^(-1/2) * (-1)
Simplifying this expression, we have:
dy/dx = (4-x)^(-1/2) - x(4-x)^(-1/2)/2
To find the critical numbers, we need to set the derivative equal to zero and solve for x:
0 = (4-x)^(-1/2) - x(4-x)^(-1/2)/2
Multiplying both sides of the equation by (4-x)^(1/2), we get:
0 = 1 - x/2
Adding x/2 to both sides of the equation, we have:
x/2 = 1
Multiplying both sides by 2, we find:
x = 2
So the critical number of the function y = x(4-x)^(1/2) is x = 2.
It seems that you obtained x = 7/2 as a potential critical number. Let's check this by substituting x = 7/2 into the derivative expression:
dy/dx = (4-x)^(-1/2) - x(4-x)^(-1/2)/2
Substituting x = 7/2, we have:
dy/dx = (4 - 7/2)^(-1/2) - (7/2)(4 - 7/2)^(-1/2)/2
Simplifying this expression, we get:
dy/dx = (1/2)^(-1/2) - (7/2)(1/2)^(-1/2)/2
Simplifying further, we obtain:
dy/dx = 2 - (7/2)(1/2)
dy/dx = 2 - 7/4
dy/dx = 8/4 - 7/4
dy/dx = 1/4
Since the derivative is not equal to zero, x = 7/2 is not a critical number of the function. The correct critical number is x = 2.
Therefore, the correct answer for the critical number is x = 2, not x = 8/3.