What are the points of inflection, if any, of #f(x)= 35x^4-36x^2 + 5x #?

Question

What are the points of inflection, if any, of #f(x)= 35x^4-36x^2 + 5x      #?

Charles Anctil · Accepted Answer

#x=+-sqrt(6/35)#

A change in a function's concavity or the sign of the second derivative—that is, a change from positive to negative or vice versa—signifies the occurrence of a point of inflection.

Using the power rule, first determine the second derivative:

#f(x)=35x^4-36x^2+5x#

#f'(x)=140x^3-72x#

#f''(x)=420x^2-72#

The sign of the second derivative could shift when the second derivative equals #0#.

#420x^2-72=0#

#420x^2=72#

#x^2=6/35#

#x=+-sqrt(6/35)#

There are two possible points of inflection at #x=-sqrt(6/35)# and #x=sqrt(6/35)#. However, they are not guaranteed to be the spot where the sign shifts, so we should plug in values surrounding these points.

To find out if the second derivative's sign really shifts, we can test the intervals around the two potential points of inflection.

When #x < -sqrt(6/35)#:

#f''(-1)=420(-1)^2-72=348#

When #-sqrt(6/35) < x < sqrt(6/35)#:

#f''(0)=420(0)^2-72=-72#

When #x > sqrt(6/35)#:

#f''(1)=420(1)^2-72=348#

Notice that at #x=-sqrt(6/35)#, the sign of the second derivative switches from positive to negative. Also, at #x=sqrt(6/35)#, the sign switches from negative to positive. Since the sign switches in both of these places, these are both points of inflection.

Charles Anctil · Answer

undefined To find the points of inflection, we need to find the second derivative of the function and then solve for the values of $ x $ where the second derivative is equal to zero or undefined.

First derivative: $ f'(x) = 140x^3 - 72x + 5 $

Second derivative: $ f''(x) = 420x^2 - 72 $

Setting $ f''(x) $ equal to zero and solving for $ x $:

$ 420x^2 - 72 = 0 $

$ 420x^2 = 72 $

$ x^2 = \frac{72}{420} $

$ x^2 = \frac{6}{35} $

$ x = \pm \sqrt{\frac{6}{35}} $

Therefore, the points of inflection are $ x = -\sqrt{\frac{6}{35}} $ and $ x = \sqrt{\frac{6}{35}} $.