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

Answer 1

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

A necessary condition for the curve #y=f(x)# to have an inflection point for #x=barx# is that:
#f''(barx) = 0#

Analyze the function's second derivative:

#f(x) = x^4/12-2x^2+15#
#f'(x) = x^3/3-4x#
#f''(x) = x^2 -4#

Based on the formula:

#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 think about the disparity:

#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.
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Answer 2

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

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.

Next, the first derivative is as follows:

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

The second derivative is thus as follows:

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

We seek out the coordinates where the second derivative vanishes, or inflection points:

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

When the first derivative vanishes at a point, it is considered a stationary point of inflection; if not, it is considered 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)#

Examining the function's graphs in relation to the first and second derivatives can be fascinating:

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

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

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 [-10, 18, 6, -10]}

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 [-10, 18, 6, 6, -10]}

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Answer 3
To find the points of inflection of \( f(x) = \frac{x^4}{12} - 2x^2 + 15 \), we first find its second derivative and then solve for \( x \) where the second derivative equals zero. After that, we determine the concavity of the function around those points. First, let's find the second derivative of \( f(x) \): \[ f(x) = \frac{x^4}{12} - 2x^2 + 15 \] \[ f'(x) = \frac{1}{3}x^3 - 4x \] \[ f''(x) = x^2 - 4 \] Now, set \( f''(x) = 0 \) to find potential points of inflection: \[ x^2 - 4 = 0 \] \[ x^2 = 4 \] \[ x = \pm 2 \] To determine the concavity around these points, we evaluate the sign of the second derivative: - For \( x < -2 \), \( f''(x) > 0 \), so the function is concave up. - For \( -2 < x < 2 \), \( f''(x) < 0 \), so the function is concave down. - For \( x > 2 \), \( f''(x) > 0 \), so the function is concave up. Therefore, the points of inflection are \( x = -2 \) and \( x = 2 \).
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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