How do you find the inflection points of the graph of the function: # f(x) = (6x)/(x^2 + 16)#?

Question

Evelyn Agard · Accepted Answer

# x = 0, f(0) = 0 " and " x = pm 4sqrt(3), f(pm 4sqrt(3)) = pm (3sqrt(3))/8 # A point of inflection can be found when the second derivative of f(x) is equal to zero i.e. #f''(x) = 0# Using quotient rule: # (df)/dx = (6(x^2+16)- 12x^2)/(x^2+16)^2 = -(6(x^2-16))/(x^2+16)^2 # # (d^2f)/dx^2 = -(12x(x^2+16)-24x(x^2-16))/(x^2+16)^3 = (12x(x^2-48))/(x^2+16)^3 = 0# # x = 0, f(0) = 0 " and " x = pm 4sqrt(3), f(pm 4sqrt(3)) = pm (3sqrt(3))/8 #

Isaiah Allen · Answer

undefined To find the inflection points of the graph of the function $ f(x) = \frac{6x}{x^2 + 16} $, follow these steps:

1. Find the second derivative of the function $ f(x) $.
2. Set the second derivative equal to zero and solve for $ x $.
3. Determine whether the solutions obtained in step 2 correspond to inflection points by analyzing the concavity of the graph.

Let's begin with step 1:

First derivative:
$$ f'(x) = \frac{d}{dx} \left( \frac{6x}{x^2 + 16} \right) $$

Using the quotient rule:
$$ f'(x) = \frac{(6)(x^2 + 16) - (6x)(2x)}{(x^2 + 16)^2} $$

Simplify:
$$ f'(x) = \frac{6x^2 + 96 - 12x^2}{(x^2 + 16)^2} $$
$$ f'(x) = \frac{-6x^2 + 96}{(x^2 + 16)^2} $$

Second derivative:
$$ f''(x) = \frac{d}{dx} \left( \frac{-6x^2 + 96}{(x^2 + 16)^2} \right) $$

Using the quotient rule:
$$ f''(x) = \frac{(x^2 + 16)^2 \cdot (-12x) - (-6x^2 + 96) \cdot 2(x^2 + 16)(2x)}{(x^2 + 16)^4} $$

Simplify:
$$ f''(x) = \frac{-12x(x^2 + 16)^2 + 24x(-6x^2 + 96)(x^2 + 16)}{(x^2 + 16)^4} $$
$$ f''(x) = \frac{-12x(x^2 + 16) + 24x(-6x^2 + 96)}{(x^2 + 16)^3} $$

$$ f''(x) = \frac{-12x^3 - 192x + 144x^3 + 2304x}{(x^2 + 16)^3} $$
$$ f''(x) = \frac{132x^3 + 2112x}{(x^2 + 16)^3} $$

Now, move on to step 2:

Set $ f''(x) = 0 $:
$$ \frac{132x^3 + 2112x}{(x^2 + 16)^3} = 0 $$

Since the numerator cannot be zero, set $ 132x^3 + 2112x = 0 $:
$$ 132x(x^2 + 16) = 0 $$

This gives us two solutions: $ x = 0 $ and $ x^2 + 16 = 0 $, which has no real solutions.

Finally, for step 3, analyze the concavity of the graph around the critical points:

Evaluate the sign of $ f''(x) $ around the critical point $ x = 0 $ to determine the concavity:
$$ f''(x) = \frac{132x^3 + 2112x}{(x^2 + 16)^3} $$

Substitute a value from each interval into $ f''(x) $ (e.g., $ x = -1 $ and $ x = 1 $):
$$ f''(-1) = \frac{132(-1)^3 + 2112(-1)}{((-1)^2 + 16)^3} $$
$$ f''(1) = \frac{132(1)^3 + 2112(1)}{(1^2 + 16)^3} $$

After calculating, you'll find that $ f''(-1) $ and $ f''(1) $ have opposite signs, indicating a change in concavity around $ x = 0 $. Hence, $ x = 0 $ is an inflection point.

Therefore, the inflection point of the graph of the function $ f(x) = \frac{6x}{x^2 + 16} $ is $ x = 0 $.