Question

Find the interval(s) on which the function f(x) = x ^ 4 - 4x ^ 3 is concave down.

Answer 1

Given the following function::

`f(x)=x^4-4x^3`

We will find the interval on which the function will be concave down

We need to find the critical points (Minimum, Maximum or Inflection )

We will find f'(x) and f''(x):

`\begin{gathered} f^(\prime)(x)=4x^3-12x^2 \\ f^(\prime)^(\prime)(x)=12x^2-24x \end{gathered}`

From f'(x) = 0 ⇒ we will get the minimum and the maximum points:

`\begin{gathered} 4x^3-12x^2=0 \\ x^2(4x-12)=0 \\ x^2=0\to x=0 \\ 4x-12=0\to x=(12)/(4)=3 \\ x=\lbrace0,3\rbrace \end{gathered}`

From, f''(x) = 0 ⇒ we will get the inflection points

`\begin{gathered} 12x^2-24x=0 \\ 12x(x-2)=0 \\ 12x=0\to x=0 \\ x-2=0\to x=2 \\ x=\lbrace0,2\rbrace \end{gathered}`

So, we can conclude the following:

There is two inflection points x = 0, x = 2

And the function has a minimum at x = 3

So, as x > 2 the function will be concave up

So, the function from x = 0 to x = 2, will be concave down

So, the answer will be: C. (0, 2)