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f(x) has an inflection point where x = -2

f(x) has an inflection point where x = 0

f(x) has an inflection point where x = 2

f(x) has a horizontal tangent where x = 0

or none of the above is true

2006-10-14 05:12:23 · 5 answers · asked by Doug 2 in Science & Mathematics Mathematics

5 answers

f'(x) = 12x^3 - 24x^2 - 48x + 96 = 12(x^3 - 2x^2 - 4x + 8)
f''(x) = 12(3x^2 - 4x - 4)

f'(x) has roots -2 and 2
f''(x) has roots 2 and -2/3

f(x) has an inflection point where x=2, but not at 0 or -2
f(x) does not have a horizontal tangent where x=0

2006-10-14 05:20:02 · answer #1 · answered by James L 5 · 0 0

f(x) has an inflection point at x=2 and -2/3 both and stn point (horizontal tang) at x=2 and -2.
Use your 1st derivitive we can find this from 12x^3-24x^2-48x+96=0 to (x-2)^2(x+2)=0 then we say stn pts are x=2 and -2
At the same time the 2nd derivative is 36x^2-48x-48 and we let it equal to 0. Therefore it means that 3x^2-4x-4=0 and (3x+2)(x-2)=0. Furthermore the points of inflexion are x=-2/3 and 2
Therefore f(x) has an inflection pt at x = 2

2006-10-14 05:33:15 · answer #2 · answered by Anonymous · 0 0

f(x) has an inflection point where x = 2 is true

2006-10-14 05:19:10 · answer #3 · answered by shamu 2 · 0 0

f(x) has an inflection point where x=2

and i have an infection point called my little brother

2006-10-14 15:35:05 · answer #4 · answered by crazyface5@sbcglobal.net 2 · 0 0

i dunno

2006-10-14 05:15:56 · answer #5 · answered by manofmusic9111 1 · 0 1

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