Question

the function ( y = f(x) ) is twice differentiable everywhere with ( f(x)=(x - 1)^{4}(3 - x)(x + 1)^{2} ). find all intervals where the function ( y = f(x) ) is concave down.
( \bigcirc(1,3) )
( \bigcirc(-infty,-1) ) and ( (3,infty) )
( \bigcirc(-infty,-1) ) and ( (1,3) )
( \bigcirc(-1,1) ) and ( (3,infty) )
( \bigcirc(3,infty) )

Explanation:

Step1: Determine the sign of \(f''(x)\)

A function \(y = f(x)\) is concave down when \(f''(x)<0\). Given \(f''(x)=(x - 1)^{4}(3 - x)(x + 1)^{2}\).
Since \((x - 1)^{4}\geq0\) for all \(x\in R\) (equality holds when \(x = 1\)) and \((x + 1)^{2}\geq0\) for all \(x\in R\) (equality holds when \(x=-1\)).
The sign of \(f''(x)\) is determined by the factor \((3 - x)\).

Step2: Solve the inequality \(f''(x)<0\)

We want to find when \((x - 1)^{4}(3 - x)(x + 1)^{2}<0\).
Since \((x - 1)^{4}\) and \((x + 1)^{2}\) are non - negative, the inequality \((x - 1)^{4}(3 - x)(x + 1)^{2}<0\) is equivalent to \(3 - x<0\) (because when \(x
eq1\) and \(x
eq - 1\), \((x - 1)^{4}>0\) and \((x + 1)^{2}>0\)).
Solving \(3 - x<0\) gives \(x>3\).
When \(x = 1\), \(f''(1)=(1 - 1)^{4}(3 - 1)(1 + 1)^{2}=0\); when \(x=-1\), \(f''(-1)=(-1 - 1)^{4}(3+1)(-1 + 1)^{2}=0\).

Answer:

\((3,\infty)\)