What are the points of inflection of #f(x)=x^{2}e^{11 x}
#?

Question

What are the points of inflection of #f(x)=x^{2}e^{11 x}
  #?

Elijah Abram · Accepted Answer

#x = +-sqrt(2/11)-2/11#.

Sufficient condition for giving points of inflexion, if any#:r f'' is not zero and f''=0. Here, #f'=2xe^(11x)+11x^2e^(11x)=e^(11x)(11x^2+2x)# #f''=e^(11x)(121x^2+44x+2)# #f'''=e^(11x)(1331x^2+726x+66)# f''=0 results in #x=+-sqrt(2/11)-2/11# and #f'''(+-sqrt(2/11)-2/11)# are not 0. So, the points of inflexion are #x =+-sqrt(2/11)-2/11#

Emily Ammerman · Answer

undefined To find the points of inflection for the function $ f(x) = x^2 e^{11x} $, we need to locate where the concavity changes. Points of inflection occur where the second derivative changes sign.

1. Find the first derivative of $ f(x) $:
   $$ f'(x) = 2x e^{11x} + x^2 (11e^{11x}) $$

2. Find the second derivative of $ f(x) $:
   $$ f''(x) = 2e^{11x} + 2x(11e^{11x}) + 22x e^{11x} $$

3. Set $ f''(x) $ equal to zero and solve for $ x $:
   $$ 2e^{11x} + 2x(11e^{11x}) + 22x e^{11x} = 0 $$
   $$ e^{11x}(2 + 22x + 22x^2) = 0 $$
   Since $ e^{11x} $ is always positive, we solve for the polynomial inside the parentheses:
   $$ 2 + 22x + 22x^2 = 0 $$

4. Solve the quadratic equation $ 22x^2 + 22x + 2 = 0 $ to find the values of $ x $.

After finding the values of $ x $, evaluate the concavity of $ f(x) $ around these points to determine whether they are points of inflection.

What are the points of inflection of #f(x)=x^{2}e^{11 x} #?