Find relative maxima and minima of
g(x)= 2x^2+ 4000/x +10
g ' (x) = 4x - 4000/x^2 + 0
= 0 for a max/min
4x = 4000/x^2
4x^3 = 4000
x^3 = 1000
x = 10
when x = 10
f(10) = 200 + 400 + 10 = 610
there is a relative min 610 at x=10
To find the relative maxima and minima of the function g(x) = 2x^2 + 4000/x +10, we need to find the critical points of the function first.
Step 1: Find the derivative of g(x) with respect to x.
g'(x) = d/dx (2x^2 + 4000/x + 10)
= 4x - 4000/x^2
Step 2: Set the derivative equal to zero and solve for x to find the critical points.
4x - 4000/x^2 = 0
4x^3 - 4000 = 0
Step 3: Solve the equation by factoring or by using the cubic formula. In this case, factoring is possible.
4(x^3 - 1000) = 0
(x^3 - 1000) = 0
Step 4: Solve for x by taking the cube root of both sides.
x^3 = 1000
x = ∛1000
x ≈ 10
So, the critical point of the function g(x) occurs at x ≈ 10.
Step 5: Determine the nature of the critical point by analyzing the second derivative.
To determine whether the critical point is a relative maximum or minimum, we need to find the second derivative of g(x).
Step 6: Find the second derivative of g(x) with respect to x.
g''(x) = d^2/dx^2 (4x - 4000/x^2)
= 4 + 8000/x^3
Step 7: Substitute the critical point x ≈ 10 into the second derivative.
g''(10) = 4 + 8000/10^3
= 4 + 8000/1000
= 4 + 8
= 12
Step 8: Analyze the second derivative to determine the nature of the critical point.
Since g''(10) = 12 is positive, this means that the second derivative is positive at x ≈ 10. Therefore, the critical point x ≈ 10 corresponds to a relative minimum of the function g(x).
Hence, the relative minimum occurs at x ≈ 10.