How do you find the critical numbers of f(x)=(x^2)(e^(11x))?

1 Answer
Feb 9, 2018

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x \ = \ 0, - 2/11.

Explanation:

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\mbox{The critical numbers} \ \ f(x) \ \ \mbox{are the solutions of:}

\qquad \qquad \qquad \qquad \ \ f'(x) \ = \ 0 \ \ \mbox{and} \ \ f'(x) \ = \ \mbox{undefined} .

\mbox{So the first step will be to calculate} \ f'(x), \mbox{and to do this, we} \ \mbox{will use the product rule as the main part of the computation.

\mbox{1) Given:} \qquad \qquad \qquad \qquad \qquad f(x) \ = \ ( x^2 ) ( e^{ 11 x } ).

\mbox{2) Product Rule:} \qquad \ \ f'(x) \ = \ ( x^2 ) [ ( e^{ 11 x } ) ]' \ + \ ( x^2 )' [ ( e^{ 11 x } ) ].

\mbox{3) Special Function Rules -- Power, Exponential Functions:}

\qquad \qquad f'(x) \ = \ ( x^2 ) [( 11 x )'( e^{ 11 x } ) ] \ + \ ( 2 x ) [( e^{ 11 x } ) ];

\qquad \qquad f'(x) \ = \ 11 ( x^2 ) ( e^{ 11 x } ) \ + \ ( 2 x ) ( e^{ 11 x } ).

\mbox{5) Simplification -- Factor Out Common Factor} \ \ e^{ 11 x } \mbox{:}

\qquad \qquad f'(x) \ = \ [ 11 ( x^2 ) \ + \ ( 2 x ) ] ( e^{ 11 x } );

\qquad \qquad f'(x) \ = \ (11 x^2 \ + \ 2 x ) e^{ 11 x }.

\mbox{4) Critical Numbers:}

\qquad \qquad \mbox{a) Solve:} \ \ f'(x) \ = \ 0.

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (11 x^2 \ + \ 2 x ) e^{ 11 x } \ = \ 0

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (11 x^2 \ + \ 2 x ) e^{ 11 x } \cdot e^{ -11 x } \ = \ 0 \cdot e^{ -11 x }

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (11 x^2 \ + \ 2 x ) e^{0 } \ = \ 0

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (11 x^2 \ + \ 2 x ) cdot 1 \ = \ 0

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad 11 x^2 \ + \ 2 x \ = \ 0

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad x (11 x \ + \ 2 ) \ = \ 0

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad x \ = \ 0 \qquad \mbox{or} \qquad x \ = \ - 2/11.

\qquad \qquad \mbox{b) Solve:} \ f'(x) = \mbox{undefined. (Where is} \ f'(x) \ \mbox{undefined ?)}

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (11 x^2 \ + \ 2 x ) e^{ 11 x } = \mbox{undefined}

\qquad \qquad \mbox{Clearly,} \ (11 x^2 \ + \ 2 x ) e^{ 11 x } \ \mbox{is defined everywhere.}

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \mbox{So no solutions here here.}

\qquad \qquad \mbox{c) Combine solutions from parts (a) and (b):}

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad x \ = \ 0, - 2/11.

\qquad \qquad \qquad \qquad \qquad \qquad \mbox{These are our critical points.}

\qquad \qquad \mbox{d) State Solution:}

\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad f(x) \ = \ ( x^2 ) ( e^{ 11 x } ).

\mbox{Critical points:} \qquad \qquad \qquad x \ = \ 0, - 2/11.