What are the critical points and also inflection points of #3e^(-2x(^2))#?

Question

Aurora Alvarado · Accepted Answer

Assuming that the function is: #f(x) = 3e^(-2x^2)#, the only critical number is: #0#, and the inflection points are: #(1/sqrt12, 3e^(-1/6))# and #(-1/sqrt12, 3e^(-1/6))#

#f(x) = 3e^(-2x^2)#

#f'(x) = 3e^(-2x^2) * (-4x) = -12xe^(-2x^2)#

#f'(x)# is never undefined and is #0# when #x=0#.

The only critical number is #0#.

#f''(x) = -12e^(-2x^2) -12x(-12xe^(-2x^2))#

# = -12e^(-2x^2)(1-12x^2)#

#f''(x) is never undefined, so the only chance it has to change signs is when it is zero.

Solving: #-12e^(-2x^2)(1-12x^2) = 0#, gives us

#x = +-1/sqrt12#

Observe that the sign of #f''(x)# is the opposite of the sign of #1-12x^2# which does, indeed, change sign at both #+1/sqrt12# and #-1/sqrt12#

#f# is an even function so we calculate: #f( +-1/sqrt12) = 3e^(-2(1/12)) = 3e^(-1/6)#

The inflection points are: #(1/sqrt12, 3e^(-1/6))# and #(-1/sqrt12, 3e^(-1/6))#

(If you'd rather, rationalize and simplify.)

Samantha Adkison · Answer

undefined To find the critical points and inflection points of $ f(x) = 3e^{-2x^2} $, we first find the first and second derivatives of the function.

The first derivative $ f'(x) $ is:

$$ f'(x) = -12xe^{-2x^2} $$

The second derivative $ f''(x) $ is:

$$ f''(x) = (24x^2 - 12)e^{-2x^2} $$

To find critical points, we set the first derivative equal to zero and solve for $ x $:

$$ -12xe^{-2x^2} = 0 $$

This equation gives us $ x = 0 $ as the only critical point.

To find inflection points, we set the second derivative equal to zero and solve for $ x $:

$$ (24x^2 - 12)e^{-2x^2} = 0 $$

This equation gives us $ x = \pm \frac{1}{\sqrt{2}} $ as the possible inflection points.

So, the critical point of the function $ f(x) = 3e^{-2x^2} $ is $ x = 0 $, and the possible inflection points are $ x = \pm \frac{1}{\sqrt{2}} $.