What are the points of inflection of #f(x)=x^2 / (x^2 + 49)

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Question

What are the points of inflection of #f(x)=x^2 / (x^2 + 49)

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Christopher Allers · Accepted Answer

The points of inflection are #(7/sqrt3,1/4)# and #(-7/sqrt3,1/4)# The points of inflections are when #f''(x)=0# Let start by calculating the derivatves, we use #f'(u/v)=(u'v-uv')/v^2# So #f'(x)=(2x(x^2+49)-x^2(2x))/(x^2+49)^2=(98x)/(x^2+49)^2# And #f''(x)=(98(x^2+49)^2-98x(x^2+49)*2*2x)/((x^2+49)^4)# #=(98(x^2+49)((x^2+49)-4x^2))/((x^2+49)^4)# #=(98(49-3x^2))/((x^2+49)^3# so #f''(x)=0# when #49-3x^2=0# that is #x=+-7/sqrt3# And #y=(49/3)/(49/3+49)=1/4# So the points are #(7/sqrt3,1/4)# and #(-7/sqrt3,1/4)#

Emily Ammerman · Answer

undefined To find the points of inflection of $ f(x) = \frac{x^2}{x^2 + 49} $, follow these steps:

1. Find the second derivative of $ f(x) $, denoted as $ f''(x) $.
2. Set $ f''(x) $ equal to zero and solve for $ x $.
3. Determine whether the solutions from step 2 correspond to points where the concavity of the function changes.

Let's begin:

1. First derivative of $ f(x) $:
$$ f'(x) = \frac{d}{dx}\left(\frac{x^2}{x^2 + 49}\right) $$
Using the quotient rule:
$$ f'(x) = \frac{(2x)(x^2 + 49) - (x^2)(2x)}{(x^2 + 49)^2} $$
$$ f'(x) = \frac{2x(x^2 + 49) - 2x^3}{(x^2 + 49)^2} $$
$$ f'(x) = \frac{98x}{(x^2 + 49)^2} $$

2. Second derivative of $ f(x) $:
$$ f''(x) = \frac{d}{dx}\left(\frac{98x}{(x^2 + 49)^2}\right) $$
Using the quotient rule again:
$$ f''(x) = \frac{(98)(x^2 + 49)^2 - 2(98x)(2x)(x^2 + 49)}{(x^2 + 49)^4} $$
$$ f''(x) = \frac{98(x^2 + 49) - 392x^2}{(x^2 + 49)^3} $$
$$ f''(x) = \frac{98x^2 + 4802 - 392x^2}{(x^2 + 49)^3} $$
$$ f''(x) = \frac{-294x^2 + 4802}{(x^2 + 49)^3} $$

3. Setting $ f''(x) $ equal to zero:
$$ \frac{-294x^2 + 4802}{(x^2 + 49)^3} = 0 $$
$$ -294x^2 + 4802 = 0 $$
$$ -294x^2 = -4802 $$
$$ x^2 = \frac{4802}{294} $$
$$ x^2 \approx 16.327 $$

Solving for $ x $:
$$ x \approx \pm 4.042 $$

Since the second derivative changes sign at these points, they correspond to the points of inflection.

Therefore, the points of inflection for $ f(x) = \frac{x^2}{x^2 + 49} $ are approximately $ x \approx -4.042 $ and $ x \approx 4.042 $.