What is the x coordinate of the absolute minimum of 6x + [12/x] when x > 0?
let y = 6x + 12/x
dy/dx = 6 - 12/x^2
= 0 for a max/min
6 = 12/x^2
x^2 = 2
x = √2, since x>0
when x = 2, y = 12 - 12/2 = 6
so the min is 6 when x = 2
I tried that answer but it wasn't right. I tried sqrt 2 and it was right.
To find the x-coordinate of the absolute minimum of the given function, we need to find the critical points of the function and check which one corresponds to the absolute minimum.
Step 1: Find the derivative of the function.
Differentiate the function 6x + [12/x] with respect to x:
f'(x) = 6 - 12/x^2.
Step 2: Find the critical points.
To find the critical point(s), we set the derivative equal to zero and solve for x:
6 - 12/x^2 = 0.
Simplifying the equation, we have:
6 = 12/x^2.
Step 3: Solve for x.
To solve for x, we can cross-multiply and then take the square root of both sides:
6x^2 = 12.
Divide both sides by 6:
x^2 = 2.
Taking the square root of both sides:
x = ±√2.
Step 4: Determine the nature of the critical point(s).
Since the original problem specifies x > 0, we only consider the positive value of x: x = √2. To determine whether x = √2 corresponds to a minimum or maximum, we can apply the second derivative test.
Differentiating the derivative from step 1:
f''(x) = 24/x^3.
Substituting x = √2 into f''(x):
f''(√2) = 24/(√2)^3 = 24/(2√2) = 12/√2 = 12√2/2 = 6√2.
Since f''(√2) = 6√2 > 0, the function has a minimum at x = √2.
Therefore, the x-coordinate of the absolute minimum of 6x + [12/x] when x > 0 is x = √2.