Critical and Inflection point help

Question

Let g(x) = e^(−6x^2)

Find and classify all critical points of g(x).

Find all inflection points of g(x).

Answer 1

This is a Gaussian function, i.e. a bell curve.  First, you should plot this in a computer or a graphing calculator. 

 

To identify the critical points, take the derivative and set it to zero!!

 

g'(x) = e^(-6x^2) * (-6*2x) = -12x*e^(-6x^2)

g'(x) = 0 at only one point, i.e. x=0

 

Therefore, the function is flat (slope = 0) at x=0

 

The value of g(0) is 1, and g(x) < 1 for all x =/= 1 (again, check on a graphing calculator).  Therefore, 0 is a global maximum.

 

Inflection points require the second derivative!!!  Use the product rule in this case.

 

g''(x) = ( [-12*e^(-6x^2)]+[-12x*(-12x)*e^(-6x^2)] ) = (-12+144*x^2)*e^(-6x^2)

 

g''(x) = 0 at -12+144*x^2 = 0   ==>   x = +/- Sqrt(1/12) = +/- 0.289

Therefore, the curvature of g(x) changes at x = Sqrt(1/12), on both sides of the y-axis (positive AND negative).

 

The value of g(+/- Sqrt(1/12)) is e^(-1/2) = 0.607

The slope of g(+/- Sqrt(1/12)) is g'(+/- Sqrt(1/12))  = -12*(+/-Sqrt(1/12))*e^(-1/2)  = -/+ Sqrt(12)*e^(-1/2) = -/+ 2.101

 

Notice the slope at the inflection points is NEGATIVE on the POSITIVE SIDE of the y-axis, and POSITIVE on the NEGATIVE SIDE of the y-axis.

 

Hope this helps.

 

Best,

Paul

Answer 2

Hi, Joan,

 

Critical points are points of a function where the slope is zero (horizontal), and so they could potentially be a maximum or minimum. To find them in this case, let's take the derivative of the function:

 

    g(x) = e^{-6x^2}

 

    g'(x) = e^{-6x^2} * (-12x)

 

We want to find the x values that would make this derivative equal to zero. There are no x values that would make the exponential term zero, so the only critical point would occur at:

 

    -12x = 0, so x=0

 

The derivative is equal to 0 at x=0, so this could be a maximum or a minimum. There are a couple of ways to determine which - let's take the second derivative, since we have to do it for the second part of the question anyway:

 

    g''(x) = e^{-6x^2} * (-12) + (-12x)*(e^{-6x^2})*(-12x)

 

    g''(x) = e^{-6x^2} * (144x² - 12)

 

Since we know that there is a critical point at x=0, let's plug in x to determine the concavity. If it comes out to be a positive number, the function is concave up at that point, meaning the function reaches a minimum. If it turns out to be a negative number, it is concave down at that point, meaning the function reaches a maximum.

 

    g''(0) = e⁰ * (0 - 12) = -12

 

At x=0, the slope is zero, and the function is concave down, so the function reaches a maximum at x=0.

 

Inflection points are points where the second derivative is equal to 0, meaning the concavity changes from up to down or vice versa. Looking back at our second derivative expression, again, the exponential term will never be equal to zero, so the only critical points exist wherever:

  

    144x² - 12 = 0

 

    144x² = 12

 

    x² = 12/144 = 1/12

 

    x = +/- √(1/12)

 

The function has inflection points (where concavity changes) at x = -√(1/12) and x = √(1/12)

 

I hope this helps!

 

 

Answer 3

I'd start with mapping out the obvious.  

 

What happens at x= 0, 1 and 2?

 

What about -1?  

 

What about 1/2 and 1/6?

 

What's special about the derivative of e^x?

 

How do you apply the chain rule to that?

 

I'm only seeing one significant point, at X = 0.