f(x)=x^4-4x^3+10
Find the inflection point(s), if there are any.
f'' = 12x^2 - 24x = 12x(x-2)
looks like inflections at x=0 and 2
To find the inflection point(s) of a function, you need to determine where the concavity changes. An inflection point occurs when the second derivative changes sign.
Let's start by finding the second derivative of the function f(x). The second derivative represents the concavity of the function.
First, find the first derivative of f(x):
f'(x) = 4x^3 - 12x^2
Now, differentiate the first derivative to find the second derivative (f''(x)):
f''(x) = 12x^2 - 24x
We have the second derivative, f''(x) = 12x^2 - 24x.
Next, set f''(x) equal to zero and solve for x to find the values where the concavity changes:
12x^2 - 24x = 0
Factor out 12x:
12x(x - 2) = 0
Now, set each factor equal to zero and solve for x:
12x = 0 or x - 2 = 0
For the first equation, divide both sides by 12:
x = 0
For the second equation, add 2 to both sides:
x = 2
So, we have two potential inflection points: x = 0 and x = 2.
To confirm whether these points are actual inflection points, we need to analyze the concavity on either side of each point. We can do this by plugging in values in the intervals (-∞, 0), (0, 2), and (2, +∞) into the second derivative, f''(x). If the sign of f''(x) changes, there is an inflection point.
Let's select a value less than 0, such as x = -1, to test the interval (-∞, 0):
f''(-1) = 12(-1)^2 - 24(-1) = 12 + 24 = 36 > 0
Since f''(-1) > 0, the concavity remains positive in the interval (-∞, 0).
Next, let's choose a value between 0 and 2, such as x = 1:
f''(1) = 12(1)^2 - 24(1) = 12 - 24 = -12 < 0
Here, f''(1) < 0, indicating a change in concavity, which means there is an inflection point at x = 1.
Lastly, let's select a value greater than 2, such as x = 3, to test the interval (2, +∞):
f''(3) = 12(3)^2 - 24(3) = 108 - 72 = 36 > 0
Since f''(3) > 0, the concavity remains positive in the interval (2, +∞).
Therefore, there is only one inflection point at x = 1.