Answer:
at the right point of inflection, near x = 0.155
Explanation:
Given a graph of a quartic function, you want to know where the second derivative becomes positive.
Derivatives
The first derivative is the slope of the curve. It will be a cubic, with zeros at approximately x = -3, x = -1, and x = +1.
The second derivative is the slope of the first derivative. It represents the concavity of the given function. To the left of about x=-2, the graph is concave upward. Between about x=-2 and x=0, the graph is concave downward, so the second derivative is negative.
To the right of about x=0, the graph is concave upward, so the second derivative is positive.
This means the second derivative changes sign from negative to positive in the vicinity of x = 0. This point on the curve where the sign of the concavity changes is called a point of inflection.
Second derivative
If we approximate the turning points of the function as x = -3, -1, and +1, then the first derivative is some scale factor times (x +3)(x +1)(x -1). Differentiating this gives the second derivative as some scale factor times 3x² +6x +1 = 3(x +1)² -4.
This has zeros at x = -1 ±√(4/3) ≈ {-2.155, 0.155}.
The rightmost point of inflection, where the second derivative becomes positive, is at about x = 0.155.
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Additional comment
The second derivative is shown in the attachment as a dashed curve. The point of inflection is marked. Its x-coordinate is the x-intercept of the second derivative curve.
We have made a guess as to the equation of the function graphed. We may be off by some vertical scale factor and the amount of vertical translation. This should not affect the x-coordinate of the inflection point.
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