Finding Local Extrema and Determining Concavity

1. Find any local extrema of the function \(f(x)=x^{3}-6x^{2}+9x+30\).
2. Determine where \(f(x)=x^{3}-6x^{2}+9x+30\) is concave up and where it is concave down.

Solution (a) The first derivative of \(f\) is \[ f^\prime (x)=3x^{2}-12x+9=3( x-1) (x-3) \]

So, \(1\) and \(3\) are critical numbers of \(f.\) Now,

\(f^\prime (x)=3( x-1) (x-3) >0\) if \(x<1\) or \(x>3\).

\(f^\prime (x) <0\) if \(1<x<3\).

So \(f\) is increasing on \(( -\infty ,1)\) and on \((3,\infty)\); \(f\) is decreasing on \((1,3)\).

At \(1,\) \(f\) has a local maximum, and at \(3\), \(f\) has a local minimum. The local maximum value is \(f( 1) = 34\); the local minimum value is \(f(3) =30\).

(b) To determine concavity, we use the second derivative: \[ f^{\prime \prime} (x)=6x-12= 6(x-2) \]

Now, we solve the inequalities \(f^{\prime \prime} (x) <0\) and \(f^{\prime \prime} (x) >0\) and use the Test for Concavity.

\(f^{\prime \prime} (x) < 0 \qquad \hbox{ if }x<2,\qquad \hbox{ so }f \hbox{ is concave down on }( -\infty ,2)\)

\(f^{\prime \prime} (x) > 0 \qquad \hbox{ if }x>2,\qquad \hbox{ so }f \hbox{ is concave up on }(2,\infty )\)