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

1 Answer
Feb 9, 2018

Answer:

# \ #

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

Explanation:

# \ #

# \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. #