What are the points of inflection, if any, of #f(x) = x^4/12 - 2x^2 + 15 #?

2 Answers
Jul 27, 2017

The function #f(x) = x^4/12-2x^2+15# has two inflection points in #x_1=-2# and #x_2=2#.

Explanation:

A necessary condition for the curve #y=f(x)# to have an inflection point for #x=barx# is that:

#f''(barx) = 0#

Evaluate the second derivative of the function:

#f(x) = x^4/12-2x^2+15#

#f'(x) = x^3/3-4x#

#f''(x) = x^2 -4#

From the equation:

#x^2 - 4 = 0#

#x^2 = 4#

#x=+-2#

we get that the function may have an inflection point in #x_1 = -2# and #x_2 = 2#.

Now consider the inequality:

#f''(x) > 0#

As #f''(x)# is a second degree polynomial with positive leading coefficient, we have that its value is negative in the interval between the roots and positive outside.

#f''(x)# then changes sign both around #x_1# and #x_2#, so these are actually inflection points.

Jul 27, 2017

There are two non-stationary points of inflection which occur at #(+-2, 25/3)#

Explanation:

We have:

# f(x) = x^4/12-2x^2+15 #

We would normally look for critical points, that is coordinates where #f'(x)=0#, however the question does not require this so it will be skipped.

The first derivative is then:

# f'(x) = x^3/3-4x #

So the second derivative is then:

# f''(x) = x^2-4 #

We look for inflection points , which are coordinates where the second derivative vanishes:

# f''(x) = 0 => x^2-4 = 0 #
# :. x^2=4 #
# :. x = +-2 #

A point of inflection is graded as a stationary point of inflection of the first derivative vanished at the point, otherwise as a non-stationary point of inflection

So when #x=-2#, we have:

# f(-2) = 16/12-8+15 = 25/3 #
# f'(-2) = -8/3+8 = 16/3 #

And when #x=2#:

# f'(2) = 16/12-8+15=25/3#
# f'(2) = 8/3-8 = 16/3#

Hence, There are two non-stationary points of inflection which occur at #(+-2, 25/3)#

It can be interesting to see the graphs of the function compared with the first and second derivatives:

The graph of the function #y=f(x)#

graph{x^4/12-2x^2+15 [-6, 6, -10, 18]}

The graph of the function #y=f'(x)#. The zeros coincide with critical points of the curve #y=f(x)# ie the turning points of #y=f(x)#

graph{x^3/3-4x [-6, 6, -10, 18]}

The graph of the function #y=f''(x)#. The zeros coincide with critical points of the curve #y=f'(x)# ie the turning points of #y=f'(x)# or the points of inflection of #y=f(x)#

graph{x^2-4 [-6, 6, -10, 18]}