What are the inflection points of #f(x) = (x^2)(e^18x)#?

Question

Christopher Agresta · Accepted Answer

You can almost tell that the #e^18# is a distractor. It doesn't matter what the exponent is because any #e^C# is just a constant.

The slope is given by the first derivative, and the concavity by the second.

Since that is the case, when the second derivative at some #x# value is positive, it's concave up and vice versa. A function with an inflection point is either concave up, concave down, or neither. Therefore, if there's somewhere that the second derivative is equal to #0#, then it must be neither concave up nor down, and so it must be an inflection point.

#d/(dx)[f(x)] = d/(dx)[e^18 x^3] = e^18 (3x^2)#

Simply from writing out #f(x)#, you can tell that you have a basic cubic function with no vertical or horizontal shifts. So, I would expect only one inflection point, and it's probably at #x = 0#.

#d/(dx)[e^18 (3x^2)] = e^18 (6x)#

Now let's just set this to #0#:

#e^18 (6x) = 0#

#color(blue)(x = 0)#

Yes, that is the only one in reality.

Samantha Adkison · Answer

undefined To find the inflection points of $ f(x) = x^2 \cdot e^{18x} $, we need to find where the second derivative changes sign. The second derivative of $ f(x) $ is $ f''(x) = (36x^2 + 36x + 2)e^{18x} $. Setting $ f''(x) = 0 $ and solving for $ x $, we get $ x = -\frac{1}{2} $ and $ x = -1 $. These are the inflection points of the function.