When does V'(x) = 0 lead to non-real solutions for V(x) = x(10-2x)(16-2x)?

[PrudensOptimus]

V(x) = x(10-2x)(16-2x)

When i V'(x) = 0... x is always a non real number... I don't know why, can someone help me thanks.

 

[ShawnD]

Originally posted by PrudensOptimus
V(x) = x(10-2x)(16-2x)

When i V'(x) = 0... x is always a non real number... I don't know why, can someone help me thanks.

V(x) = (10x-2x^2)(16-2x)
V(x) = 160x - 20x^2 - 32x^2 + 4x^3
V(x) = 160x - 52x^2 + 4x^3
V'(x) = 160 - 104x + 12x^2
0 = 160 - 104x + 12x^2 I don't feel like doing quadratic equation so I just graphed it
x = 2, x = 6.6666666 (which is 20/3)

How did you go about getting your answers?

 

[M98Ranger]

When V'(x) = 0, it means that the derivative of the function V(x) is equal to zero. In other words, the slope of the function at that particular point is zero. This can happen at multiple points on a function, including at non-real numbers.

In this specific case, the function V(x) = x(10-2x)(16-2x) has multiple critical points where the derivative is equal to zero. These points are x = 0, x = 5, and x = 8. Therefore, when V'(x) = 0, x can take on any of these values, including non-real numbers.

To understand why this is the case, we need to look at the graph of the function. The graph of V(x) is a parabola that opens downwards, with its vertex at the point (5, 400). This means that the function has a maximum value of 400 at x = 5, and it decreases on either side of this point.

At x = 0 and x = 8, the function also has a slope of zero, but these points are not real numbers because they lie outside the domain of the function. This is because when x = 0, the function is undefined (you cannot divide by zero), and when x = 8, the function has a negative value, which is not possible for a volume.

Therefore, when V'(x) = 0, x can take on any of these values, including non-real numbers, because they are critical points of the function where the derivative is equal to zero. It is important to note that non-real numbers are still valid solutions in mathematics and can have real-world applications. I hope this helps clarify why x can be a non-real number in this situation.