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Question 999753: f(x) = x√(1-x^2)
f'(x) = (1-2x^2)/√(1-x^2)
f"(x) = (2x^3-3x)/((1-x^2)^(3/2))
now to find the inflection pts I need to set the 2nd derivative to zero and find what x equals. However, since it's a rational function I dismiss the denominator, right? because it cannot be divided by zero.
If so I get:
2x^3-3x = 0
2x^3 = 3x
x^2 = 3x
x = ± √(3/2)
technically isn't these two the inflection points?
When I plug it into my program it tells me this is wrong.
Any ideas?
Thanks
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! f(x) = x√(1-x^2)
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f'(x) = x*(-2x/sqrt(1-x^2) + sqrt(1-x^2)*1
f'(x) = [-2x^2 + (1-x^2)]/sqrt(1-x^2)
f'(x) = [-3x^2+1]/sqrt(1-x^2)
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f''(x) = [sqrt(1-x^2)*(-6x) - [-3x^2+1]*(-2x)/sqrt(1-x^2)] all over (1-x^2)
Simplify that then solve the numerator for zero.
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Cheers,
Stan H.
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now to find the inflection pts I need to set the 2nd derivative to zero and find what x equals. However, since it's a rational function I dismiss the denominator, right? because it cannot be divided by zero.
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