Given the following function::
[tex]f(x)=x^4-4x^3[/tex]
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):
[tex]\begin{gathered} f^{\prime}(x)=4x^3-12x^2 \\ f^{\prime}^{\prime}(x)=12x^2-24x \end{gathered}[/tex]
From f'(x) = 0 ⇒ we will get the minimum and the maximum points:
[tex]\begin{gathered} 4x^3-12x^2=0 \\ x^2(4x-12)=0 \\ x^2=0\to x=0 \\ 4x-12=0\to x=\frac{12}{4}=3 \\ x=\lbrace0,3\rbrace \end{gathered}[/tex]
From, f''(x) = 0 ⇒ we will get the inflection points
[tex]\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}[/tex]
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)
