How do you find local maximum value of f using the first and second derivative tests: #f(x) = e^x(x^2+2x+1)#?

Question

How do you find local maximum value of f using the first and second derivative tests:  #f(x) = e^x(x^2+2x+1)#?

Christopher Allers · Accepted Answer

#f_max = f(-3) = 4e^(-3) approx 0.19914827347#

#f(x)= e^x(x^2+2x+1)#

For a local maximum or minimum of #f(x), f'(x) =0#

Applying the product rule and the standard differential of #e^x#

#f'(x) = e^x(2x+2) + e^x(x^2+2x+1)#

#= e^x(x^2+4x+3)#

Setting #f'(x) =0#

#e^x(x^2+4x+3) =0#

Since #e^x >0 forall x#

#x^2+4x+3 =0#

#(x+3)(x+1) = 0 -> x= -3 or -1#

To test for a maximum or minimum we need to test the sign of #f''(x)# at the turning points.

Applying the product rule and the standard differential of #e^x# again.

#f''(x) = e^x(2x+4) + e^x(x^2+4x+3)#

#= e^x(x^2+6x+7)#

#f''(-3) = e^(-3)(9-18+7) = -2e^(-3) <0 -> f(-3)# is a local maximum

#f''(-1) =e^(-1)(1-6+7) = 2e^(-1)>0 -> f(-1)# is a local minimum

We are asked to find the local maximum.

#f_"max" = f(-3) = e^(-3)(9-6+1) = 4e^(-3) approx 0.19914827347#

The local extrema can be seen on the graph of #f(x)# below. graph{e^x(x^2+2x+1) [-4.783, 0.086, -1.156, 1.277]}

Christopher Allers · Answer

undefined To find the local maximum value of $ f(x) = e^x(x^2 + 2x + 1) $ using the first and second derivative tests:

1. Find the first derivative $ f'(x) $ by applying the product rule.
   $ f'(x) = e^x(x^2 + 2x + 1) + e^x(2x + 2) $

2. Simplify $ f'(x) $ to get:
   $ f'(x) = e^x(x^2 + 2x + 1 + 2x + 2) $
   $ f'(x) = e^x(x^2 + 4x + 3) $

3. Set $ f'(x) $ equal to zero to find critical points:
   $ e^x(x^2 + 4x + 3) = 0 $
   This implies $ x^2 + 4x + 3 = 0 $

4. Solve the quadratic equation $ x^2 + 4x + 3 = 0 $ to find the critical points.
   The solutions are $ x = -1 $ and $ x = -3 $.

5. Now, find the second derivative $ f''(x) $ by differentiating $ f'(x) $.
   $ f''(x) = e^x(x^2 + 4x + 3) + e^x(2x + 4) $

6. Simplify $ f''(x) $ to get:
   $ f''(x) = e^x(x^2 + 4x + 3 + 2x + 4) $
   $ f''(x) = e^x(x^2 + 6x + 7) $

7. Now, evaluate $ f''(x) $ at each critical point:
   $ f''(-1) = e^{-1}((-1)^2 + 6(-1) + 7) $
   $ f''(-1) = e^{-1}(1 - 6 + 7) $
   $ f''(-1) = e^{-1}(2) > 0 $, indicating a local minimum at $ x = -1 $.

$ f''(-3) = e^{-3}((-3)^2 + 6(-3) + 7) $
   $ f''(-3) = e^{-3}(9 - 18 + 7) $
   $ f''(-3) = e^{-3}(-2) < 0 $, indicating a local maximum at $ x = -3 $.

8. Therefore, the local maximum value of $ f(x) $ occurs at $ x = -3 $. To find the corresponding $ y $-value, plug $ x = -3 $ into the original function:
   $ f(-3) = e^{-3}((-3)^2 + 2(-3) + 1) $
   $ f(-3) = e^{-3}(9 - 6 + 1) $
   $ f(-3) = e^{-3}(4) $

So, the local maximum value of $ f(x) $ is $ 4e^{-3} $.