The function 3x^(3)+6x^(2)+6x+10 is maxima, minima or point of inflection?

1 Answer
Darshan Senthil
Feb 12, 2018
  • No mins or maxes
  • Point of Inflection at #x = -2/3#.

graph{3x^3 + 6x^2 + 6x + 10 [-10, 10, -10, 20]}

Explanation:

Mins and Maxes

For a given #x#-value (let's call it #c#) to be a max or min for a given function, it has to satisfy the following:

#f'(c) = 0# or undefined.

These values of #c# are also called your critical points.

Note: Not all critical points are max/mins, but all max/mins are are critical points

So, let's find these for your function:

#f'(x) = 0#

#=> d/dx(3x^3 + 6x^2 + 6x + 10) = 0#

#=> 9x^2 + 12x + 6 = 0#

This doesn't factor, so let's try quadratic formula:

#x = (-12 +- sqrt(12^2 - 4(9)(6)))/(2(9))#

#=> (-12 +-sqrt(-72))/18#

...and we can stop right there. As you can see, we end up having a negative number under the square root. Hence, there are no real critical points for this function.

-

Points of Inflection

Now, let's find points of inflection. These are points where the graph has a change in concavity (or curvature). For a point (call it #c#) to be a point of inflection, it must satisfy the following:

#f''(c) = 0#.

Note: Not all such points are points of inflection, but all points of inflection must satisfy this.

So let's find these:

#f''(x) = 0#

#=> d/dx(d/dx(3x^3 + 6x^2 +6x + 10)) = 0#

#=> d/dx(9x^2 + 12x + 6 = 0)#

#=> 18x + 12 = 0#

#=> x = -12/18 = -2/3#

Now, we need to check if this is in fact a point of inflection. So we'll need to verify that #f''(x)# does in fact switch sign at #x = -2/3#.

So let's test values to the right & left of #x = -2/3#:

Right:
#x = 0#
#f''(0) = 12#

Left:
#x = -1#
#f''(-1) = -6#

We don't care as much what the actual values are, but as we can clearly see, there's a positive number to the right of #x = -2/3#, and a negative number to the left of #x = -2/3#. Hence, it is indeed a point of inflection.

To summarize, #f(x)# has no critical points (or mins or maxes), but it does have a point of inflection at #x = -2/3#.

Let's take a look at the graph of #f(x)# and see what these results mean:

graph{3x^3 + 6x^2 + 6x + 10 [-10, 10, -10, 20]}

This graph is increasing everywhere, so it doesn't have any place where the derivative = 0. However, it does go from curved down (concave down) to curved up (concave up) at #x = -2/3#.

Hope that helped :)

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