How do you find the inflection points of the graph of the function: # f(x)= x^3 - 12x#?

Question

How do you find the inflection points of the graph of the function: # f(x)= x^3  -  12x#?

Aurora Alvarado · Accepted Answer

At x = 0, there is a inflection.

y = #x^3# - 12x

#dy/dx# = 3#x^2# - 12
#d^2/dx^2#=6x

Set #d^2/dx^2#=0 => 6x = 0
Take a few values on either side. The co-ordinates -

x y
-3 9
-2 16
-1 11
0 0
1 -11
2 -16
3 -9
The graph

Emma Aldridge · Answer

undefined To find the inflection points of the function $ f(x) = x^3 - 12x $, you need to 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 the $ x $ values found in step 2 where the concavity changes. Now, let's go through each step: 1. The first derivative of $ f(x) $ is $ f'(x) = 3x^2 - 12 $. 2. The second derivative of $ f(x) $ is $ f''(x) = 6x $. 3. Set $ f''(x) = 0 $ and solve for $ x $: $ 6x = 0 $ $ x = 0 $ So, the critical point is $ x = 0 $. 4. Determine the concavity of $ f(x) $ around $ x = 0 $: - For $ x < 0 $, $ f''(x) < 0 $, indicating concave down. - For $ x > 0 $, $ f''(x) > 0 $, indicating concave up. Therefore, the function changes concavity at $ x = 0 $. 5. Hence, $ x = 0 $ is the inflection point of the graph of the function $ f(x) = x^3 - 12x $.