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

What are the points of inflection of #f(x)=x^{2}e^{11 -x}
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Emilia Aguilar · Accepted Answer

#f(x)=x^2e^(11−x)# #f'(x)=2xe^(11−x)+x^2e^(11−x)*-1# #f'(x)=e^(11−x)[2x-x^2]# #f''(x)=-e^(11−x)(2x-x^2)+e^(11−x)(2-2x)# #f''(x)=e^(11−x)[-2x+x^2+2-2x]# #f''(x)=e^(11−x)[x^2-4x+2]# #e^(11−x)>0quad# That means we don't care about that. BUT: #x^2-4x+2=0# #x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)# #x_(1,2)=(4+-sqrt(4^2-4*1*2))/(2*1)# #x_(1,2)=(4+-sqrt(16-8))/(2)=(2*2+-sqrt(2^2*2))/(2)# #x_(1,2)=(cancel2*2+-cancel2sqrt(2))/(cancel2)=2+-sqrt(2)# #x_1=2-sqrt(2)# #x_2=2+sqrt(2)#

Charles Anctil · Answer

undefined To find the points of inflection of $ f(x) = x^2 e^{11 - x} $, we first need to find the second derivative, $ f''(x) $, and then determine where it equals zero and changes sign.

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

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

3. Simplify the second derivative:
$ f''(x) = 2e^{11 - x} - 2xe^{11 - x} + 2e^{11 - x} + x^2e^{11 - x} $
$ f''(x) = (x^2 - 2x + 2)e^{11 - x} $

4. Set $ f''(x) $ equal to zero and solve for $ x $:
$ (x^2 - 2x + 2)e^{11 - x} = 0 $
$ x^2 - 2x + 2 = 0 $

5. Solve the quadratic equation:
Using the quadratic formula, $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $, where $ a = 1 $, $ b = -2 $, and $ c = 2 $:
$ x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(2)}}{2(1)} $
$ x = \frac{2 \pm \sqrt{4 - 8}}{2} $
$ x = \frac{2 \pm \sqrt{-4}}{2} $

Since the discriminant is negative, there are no real solutions for $ x $, hence, no points of inflection for the given function.

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