If f ''(x) = x3(x−1)3(x+5)2, find the x-values for all the points of inflection. (Enter your answers as a comma separated list.)
Since at the points of inflection of any function f(x), f '' (x) = 0
and you are given f '' (x) already in factored form, this is quite simple.
All you have to do it set each of the factors equal to zero and solve
f '' (x) = x^3(x−1)^3(x+5)^2
x^3 = 0 ---> x = 0
(x-1)^3 = 0
x-1 = 0 ----> x = 1
(x+5)^2 = 0 ---> x = -5
Answer in the way required.
so at inflection, f" is zero
I see a triple zero a x=0, triple zero at 1, and a double at x-5
To find the x-values for all the points of inflection, we need to first determine where the concavity of the function changes.
The concavity changes when the second derivative, f''(x), changes signs. In this case, f''(x) = x^3(x-1)^3(x+5)^2.
So, we need to find the values of x for which f''(x) = 0 or is undefined.
Let's start by setting f''(x) = 0:
x^3(x-1)^3(x+5)^2 = 0
We can separate this equation into three factors:
x^3 = 0, (x-1)^3 = 0, and (x+5)^2 = 0.
For x^3 = 0, the only solution is x = 0.
For (x-1)^3 = 0, the only solution is x = 1.
For (x+5)^2 = 0, the only solution is x = -5.
Now, let's check if f''(x) is undefined at any point.
f''(x) is undefined if any of the factors x^3, (x-1)^3, or (x+5)^2 are equal to 0.
However, we have already considered these cases when solving for x = 0, x = 1, and x = -5.
So, the x-values for all the points of inflection are x = 0, x = 1, and x = -5.
Therefore, the answer is {0, 1, -5}.