Finding critical points of a single variable function

This part continues from where maximum and minimum points of a function left off. Now we deal with the specific case of [math]\displaystyle{ f''(x) = 0 }[/math]. In the same way the derivative shows us whether a function is crescent or decrescent in a certain interval, the second derivative does the same for the derivative itself. With the second derivative we know whether a function's rate of change is increasing or decreasing over time, which graphically means that the function's concavity (its curvature) is either upwards or downwards. Let's inspect the graph of a cubic and its second derivative:

[math]\displaystyle{ ]a, \ 0[ }[/math] we have that [math]\displaystyle{ f''(x) \lt 0 }[/math]. The second derivative is crescent, its sign is negative and the graph of [math]\displaystyle{ f }[/math] is a downwards parabola.

[math]\displaystyle{ ]0, \ b[ }[/math] we have that [math]\displaystyle{ f''(x) \gt 0 }[/math]. The second derivative is crescent, its sign is positive and the graph of [math]\displaystyle{ f }[/math] is an upwards parabola.

At the origin [math]\displaystyle{ f''(x) = 0 }[/math]. When [math]\displaystyle{ f'(x) = 0 }[/math] it means that we have a horizontal tangent line at that point. When [math]\displaystyle{ f''(x) = 0 }[/math] it means that the derivative has changed from crescent to decrescent or vice-versa. It's called an inflection point, the point where [math]\displaystyle{ f }[/math] inverts its rate of change from positive to negative or vice-versa. Which is exactly what we see in the graph.

The function's concavity tells us whether the function is increasing | decreasing faster or slower over time. If concavity is up, the rate of change changes faster over time. If it's down, the rate of change changes slower over time. Note that when we have concavity up or down in a certain interval, there can't be any inflection points in the interval, except for the extremes of it.

Careful! In the same way the derivative's root may or may not coincide with a local maximum or minimum, the second derivative's root may or may not coincide with the root of a function.

If [math]\displaystyle{ f'' }[/math] exists in an open interval [math]\displaystyle{ I }[/math], then:

[math]\displaystyle{ f''(x) \gt 0 }[/math] in [math]\displaystyle{ I }[/math] means that the concavity of [math]\displaystyle{ f }[/math] is up in that interval.
[math]\displaystyle{ f''(x) \lt 0 }[/math] in [math]\displaystyle{ I }[/math] means that the concavity of [math]\displaystyle{ f }[/math] is down in that interval.

Note: we are already excluding linear functions, because the derivative of a linear function is always a constant.

The graphical idea of the proof is quite simple. We have to use the Mean Value Theorem and the linear approximation. When the concavity is up, the tangent line is below the function. When the concavity is down, the tangent line is above the function. In other words, what we have to prove is that the function, is above its linear approximation.

Let [math]\displaystyle{ c }[/math] be a number that is in the interval [math]\displaystyle{ I }[/math]. For all other values we have to prove that [math]\displaystyle{ f(x) \gt T(x) }[/math], where [math]\displaystyle{ T(x) = f(c) - f'(x)(x - c) }[/math]. Remember that the point where the tangent line intersects the function, then [math]\displaystyle{ f(c) = T(c) }[/math] exactly there.

Consider that the difference between the linear approximation and the function is another function:

[math]\displaystyle{ g(c) = f(x) - T(x) }[/math], for all [math]\displaystyle{ x \in I }[/math]. For the previous inequality to be true, we have that [math]\displaystyle{ g(x) \gt 0 }[/math] must be true in that interval: