Asked by NEED Help
f(x)=ln(x^4+27)
find the inflection points?
Answers
Answered by
Reiny
f'(x) = 4x^3/(x^4 + 27)
f''(x) = (12x^2(x^4+27) - 4x^3(4x^3))/(x^4+27)^2
= 4x^2(81-x^4)/(x^4+27)^2 = 0 for inflection points
4x^2 = 0 or 81-x^4 = 0
x = 0 or (9-x^2)(9+x)^2 = 0
x = 0 or (3+x)(3-x)(9+x)^2 = 0
x = 0 or x = -3 or x = 3
So we know the x values of our 3 points of inflection, sub into the original equation to find the matching y values.
f''(x) = (12x^2(x^4+27) - 4x^3(4x^3))/(x^4+27)^2
= 4x^2(81-x^4)/(x^4+27)^2 = 0 for inflection points
4x^2 = 0 or 81-x^4 = 0
x = 0 or (9-x^2)(9+x)^2 = 0
x = 0 or (3+x)(3-x)(9+x)^2 = 0
x = 0 or x = -3 or x = 3
So we know the x values of our 3 points of inflection, sub into the original equation to find the matching y values.
Answered by
drwls
The inflection points are where the second derivative is zero.
f'(x) = 3x^2/(x^4 +27)
f''(x) = [(x^4+27)(6x) - 12x^5]/(x^4 +27)^2
f''(x) = 0 where the denominator is zero.
f'(x) = 3x^2/(x^4 +27)
f''(x) = [(x^4+27)(6x) - 12x^5]/(x^4 +27)^2
f''(x) = 0 where the denominator is zero.
Answered by
NEED Help
i am not getting the ans i mean it says you have to find smaller and larger values.
i already know that one smaller is -3
and one larger value is 3 but the other two i am not getting :(
i already know that one smaller is -3
and one larger value is 3 but the other two i am not getting :(
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