What are the points of inflection of #f(x)=2 / (x^2 - 9)

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Question

What are the points of inflection of #f(x)=2 / (x^2 - 9)
 
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Isabella Ackerman · Accepted Answer

There are no points of inflection for the graph of this function.

#f(x) = 2/(x^2-9)# has domain all reals except #3#, #-3#. #f''(x) = (12(x^3+3))/(x^2-9)^3# #f''# changes sign, so concavity changes, at #x=-3# and at #x=3#. A point on the graph where the concavoty changes is called an inflection point. The graph of this function does not have points with #x=3# or #x=-3#. So it has no inflection points.

Scarlett Allison · Answer

undefined To find the points of inflection of $ f(x) = \frac{2}{x^2 - 9} $, we need to first find the second derivative of the function. Then, we set the second derivative equal to zero and solve for the values of $ x $ to determine the points of inflection.

First, let's find the first derivative of $ f(x) $:
$$ f'(x) = \frac{d}{dx} \left( \frac{2}{x^2 - 9} \right) $$

$$ f'(x) = \frac{-4x}{(x^2 - 9)^2} $$

Now, let's find the second derivative:
$$ f''(x) = \frac{d}{dx} \left( \frac{-4x}{(x^2 - 9)^2} \right) $$

$$ f''(x) = \frac{-4(x^2 - 9)^2 - 2(-4x)(2x(x^2 - 9))}{(x^2 - 9)^4} $$

$$ f''(x) = \frac{-4(x^4 - 18x^2 + 81) + 16x^2(x^2 - 9)}{(x^2 - 9)^4} $$

$$ f''(x) = \frac{-4x^4 + 72x^2 - 324 + 16x^4 - 144x^2}{(x^2 - 9)^4} $$

$$ f''(x) = \frac{12x^4 - 72x^2 - 324}{(x^2 - 9)^4} $$

To find the points of inflection, we set the second derivative equal to zero and solve for $ x $:
$$ 12x^4 - 72x^2 - 324 = 0 $$

$$ 3x^4 - 18x^2 - 81 = 0 $$

$$ (x^2 - 9)(3x^2 + 9) = 0 $$

$$ x^2 - 9 = 0 \quad \text{(Ignoring the trivial solution)} $$

$$ x^2 = 9 $$

$$ x = \pm 3 $$

Therefore, the points of inflection of $ f(x) = \frac{2}{x^2 - 9} $ are $ x = -3 $ and $ x = 3 $.

What are the points of inflection of #f(x)=2 / (x^2 - 9) #?