Question

suppose ( f^{prime prime} ) is continuous on ( (-infty, infty) ).
(a) if ( f^{prime}(-3)=0 ) and ( f^{prime prime}(-3)=-5 ), what can you say about ( f ) ?
( \bigcirc ) at ( x=-3, f ) has a local maximum.
( \bigcirc ) at ( x=-3, f ) has a local minimum.
( \bigcirc ) at ( x=-3, f ) has neither a maximum nor a minimum.
( \bigcirc ) more information is needed to determine if ( f ) has a maximum or minimum at ( x=-3 ).
(b) if ( f^{prime}(-1)=0 ) and ( f^{prime prime}(-1)=0 ), what can you say about ( f ) ?
( \bigcirc ) at ( x=-1, f ) has a local maximum.
( \bigcirc ) at ( x=-1, f ) has a local minimum.
( \bigcirc ) at ( x=-1, f ) has neither a maximum nor a minimum.
( \bigcirc ) more information is needed to determine if ( f ) has a maximum or minimum at ( x=-1 ).

Explanation:

Part (a)

• Recall the second - derivative test: If \(f^{\prime}(c)=0\) and \(f^{\prime\prime}(c)>0\), then \(f(x)\) has a local minimum at \(x = c\). If \(f^{\prime}(c)=0\) and \(f^{\prime\prime}(c)<0\), then \(f(x)\) has a local maximum at \(x = c\).
• Given \(f^{\prime}(-3)=0\) and \(f^{\prime\prime}(-3)=-5<0\). By the second - derivative test, at \(x=-3\), \(f\) has a local maximum.

Part (b)

• The second - derivative test is inconclusive when \(f^{\prime}(c) = 0\) and \(f^{\prime\prime}(c)=0\).
• For example, consider \(y=x^{3}\), \(y^{\prime}=3x^{2}\), \(y^{\prime\prime}=6x\). At \(x = 0\), \(y^{\prime}(0)=0\) and \(y^{\prime\prime}(0)=0\), and \(y=x^{3}\) has neither a local maximum nor a local minimum at \(x = 0\).
• Consider \(y=x^{4}\), \(y^{\prime}=4x^{3}\), \(y^{\prime\prime}=12x^{2}\). At \(x = 0\), \(y^{\prime}(0)=0\) and \(y^{\prime\prime}(0)=0\), and \(y=x^{4}\) has a local minimum at \(x = 0\).
• Consider \(y=-x^{4}\), \(y^{\prime}=-4x^{3}\), \(y^{\prime\prime}=-12x^{2}\). At \(x = 0\), \(y^{\prime}(0)=0\) and \(y^{\prime\prime}(0)=0\), and \(y =-x^{4}\) has a local maximum at \(x = 0\).

Answer:

(a) At \(x=-3\), \(f\) has a local maximum.
(b) More information is needed to determine if \(f\) has a maximum or minimum at \(x=-1\).