- #1

PrudensOptimus

- 641

- 0

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

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- Thread starter PrudensOptimus
- Start date

- #1

PrudensOptimus

- 641

- 0

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

- #2

ShawnD

Science Advisor

- 712

- 2

Originally posted by PrudensOptimus

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?

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