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Pre calculus question? Please help?

+1 vote
Find all points of inflection of the graph of the function f(x) = x^3 - 12x.
asked Jan 11, 2013 in PRECALCULUS by linda Scholar

1 Answer

+1 vote

f(x) = x^3-12x

If f(x) = x^3 - 12x

[ Note : (d/dx)(x^3) = 3x^2  ,  (d/dx)x = 1 ]

Simplify

f'(x) = 3( x^2) - 12 (1)

f'(x) = 3( x^2) - 12

Note : (d/dx)(x^2) = 2x  ,  (d/dx)(const.) = 0 ]

f''(x) = 3[2x] - 0

f''(x) = 3[2x]

f''(x) = 6x

solve f'(x) = 0 , f''(x) = 0 to get the point of inflection

f'(x) = 3( x^2) - 12 = 0

⇒3( x^2) - 12 = 0

Add 12 to each side.

⇒ 3( x^2) - 12 + 12 = 12

Simplify

⇒ 3( x^2) = 12

Divide each side by 3

⇒ 3( x^2) / 3 = 12 / 3

Simplify

⇒ ( x^2 ) = 4

⇒ Root apply to each side

⇒√( x^2 ) = √4

⇒√( x^2 ) = √2^2

Cacel to square and root.   √( A^2 ) = A

⇒x = 2

⇒ f''(x) = 0

⇒ f''(x) = 6x = 0

⇒6x = 0

Divide each side by 6

⇒6x / 6 = 0 / 6

Simplify

x = 0

The roots are 2 and 0

f(x) = x^3-12x

substitute x = 2 in the equation

f(2) = (2)^3 - 12(2)

Simplify.

f(2) = 8 - 24

f(2) = - 16

f(x) = x^3-12x

substitute x = 0 in the equation

f(0) = 0^3 - 12(0)

Simplify

f(0) = 0

There fore

The points of inflection are (2 , -16 ) and (0 , 0)

answered Jan 11, 2013 by richardson Scholar

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