Estimate the $x$-coordinate of the point where $g''(x)=0$

Question

I know the only point where $g''(x)=0$ will occur when the line turns into a straight line. How do I find where that happens without an equation of the line and just a graph?

Answer 1

Hint: Look for where the concavity changes (in this case, where does the graph stop being concave down and begins to be concave up?). When $g''(x_0)=0$, $x_0$ is an inflection point.

Answer 2

I assume we are looking at the graph of $g(x)$ here, NOT the graph of $g''(x)$ so I will continue my answer based off of that assumption.

When it comes to the graph of any continuous function $g(x)$, we are able to make estimates about $g'(x)$ and $g''(x)$, the first and second derivatives of $g$ with respect to $x$, even without an equation.

The graphical meaning of the first derivative $g'(x)$ talks about the slope of the original function $g(x)$. When $g'(x) = 0$, then the original graph, the graph of $g(x)$, will be changing direction from increasing to decreasing. For this particular case, see that the graph is always increasing, so there is no $x$ such that $g'(x) = 0$

The second derivative $g''(x)$ talks about the concavity of the original graph. That is, does the graph look like it is opened up or opened down? A very simple example would be to look at $y=x^2$ and $y=-x^2$. The first equation is an example of positive concavity and the second example is one for negative concavity. A third example, $y=x^3$, includes both, as it has positive concavity to the right of $x=0$ and negative concavity to the left of $x=0$. For this specific example, $x=0$ is what is known as a inflection point.

Getting back to your example, try and see where the graph starts going from being opened down (concave down) to being opened up (concave up). At that particular value of $x$ is where $g''(x) = 0$

Answer 3

You graphed the cubic

$$ y= (x-3)(x^2+3x +32)/48 \tag1 $$

passing through

$$(0,2),(-3,-4),(5,3) $$

When above points are plugged into a quadratic of unknown coefficients you get (1).

Now successive derivatives are

$$ y^{'}= (3x^2+23)/48$$

$$ y^{''}=\frac{x}{8}$$

When above second derivative vanishes, i.e., when (here cubic) a curve is locally straight you have an inflection point always. In this case

$$ (x=0, y=-2 ) $$

is the inflection point on y-axis in your plot/graph.