relative maxima and relative minima
2x^2+ 4000/x +10
y' = 2(x^2+20x-2000)/(x+10)^2
so, we just need to find where
x^2+20x-2000 = 0
To find the relative maxima and relative minima of a function, you need to find the critical points and determine whether they correspond to a maximum or minimum.
First, let's find the derivative of the function:
f(x) = 2x^2 + 4000/x + 10
f'(x) = 4x - 4000/x^2
Set f'(x) = 0 to find the critical points:
4x - 4000/x^2 = 0
Multiplying throughout by x^2, we get:
4x^3 - 4000 = 0
Now, solve for x:
4x^3 = 4000
x^3 = 1000
x = ∛1000
x = 10
So, the critical point is x = 10. Now, let's determine whether it corresponds to a relative maximum or relative minimum.
To do this, we can use the second derivative test. Find the second derivative of the function:
f''(x) = 4 + 8000/x^3
Evaluate f''(x) at the critical point x = 10:
f''(10) = 4 + 8000/10^3 = 4 + 800 = 804
Since f''(10) is positive, it implies that the function has a relative minimum at x = 10.
Therefore, the function f(x) = 2x^2 + 4000/x + 10 has a relative minimum at x = 10.