What are the points of inflection, if any, of #f(x) = x^5/20 - 5x^3 + 5 #?

Question

What are the points of inflection, if any, of #f(x) = x^5/20 - 5x^3 + 5    #?

Samantha Adkison · Accepted Answer

The points of inflection are #x= -sqrt30, 0, and sqrt(30)#.

#f(x) = (x^5/20) - 5x^3 + 5# First, you find the second derivative and set it equal to 0. Points of inflection are places where a function's second derivative switches signs and is equal to zero. # f'(x) = (5x^4/20) - 15x^2 = (x^4/4) -15x^2 # #f''(x) = (4x^3/4) - 30x = x^3 - 30x # #x^3 - 30x = 0# Factor out the x. #x(x^2 - 30) = 0# so #x=0# and #x = +- sqrt(30) # To check if these are all inflection points, you check if the signs change around them. I plugged in #x=-6, -1, 1#, and #6#, but you can do this with any points outside and between the three answers. The signs changed around all 3 points, so all 3 are points of inflection.

Luke Alvarez · Answer

undefined To find the points of inflection of $ f(x) = \frac{x^5}{20} - 5x^3 + 5 $, we first need to find its second derivative and then identify where it equals zero.

First, find the first derivative $ f'(x) $ and then the second derivative $ f''(x) $:

$$ f'(x) = \frac{1}{20} \cdot 5x^4 - 15x^2 $$
$$ f''(x) = \frac{1}{20} \cdot 20x^3 - 30x $$

Now, set the second derivative equal to zero and solve for $ x $:

$$ f''(x) = 0 $$
$$ 20x^3 - 30x = 0 $$
$$ 10x(2x^2 - 3) = 0 $$
$$ x = 0, \sqrt{\frac{3}{2}}, -\sqrt{\frac{3}{2}} $$

These are the potential points of inflection. To determine if they are indeed points of inflection, we analyze the sign of the second derivative around each point.

- At $ x = 0 $:
  - Test point to the left: $ f''(-1) = (-1)^3 - 30(-1) = -29 $ (negative)
  - Test point to the right: $ f''(1) = 1^3 - 30(1) = -29 $ (negative)
  - Since the second derivative changes sign around $ x = 0 $, it's a point of inflection.

- At $ x = \sqrt{\frac{3}{2}} $:
  - Test point to the left: $ f''(1) = 20 - 30\sqrt{\frac{3}{2}} $ (positive)
  - Test point to the right: $ f''(2) = 160 - 60\sqrt{\frac{3}{2}} $ (positive)
  - Since the second derivative doesn't change sign around $ x = \sqrt{\frac{3}{2}} $, it's not a point of inflection.

- At $ x = -\sqrt{\frac{3}{2}} $:
  - Test point to the left: $ f''(-2) = -160 + 60\sqrt{\frac{3}{2}} $ (negative)
  - Test point to the right: $ f''(-1) = -20 + 30\sqrt{\frac{3}{2}} $ (negative)
  - Since the second derivative doesn't change sign around $ x = -\sqrt{\frac{3}{2}} $, it's not a point of inflection.

So, the only point of inflection is $ x = 0 $.