Critical point

  • #1
PrudensOptimus
641
0
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.
 

Answers and Replies

  • #2
ShawnD
Science Advisor
712
2
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?
 

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