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 positive to negative or vice-versa. It's called an inflection point. Some exercises try to trick us because that point may have the derivative changing from crescent to decrescent, which is not the same behaviour as changing its sign! This happens with polynomials of degree 4 or greater for example.

The function's concavity tells us whether the function is increasing | decreasing faster or slower over time. If concavity is up and the function is crescent in that interval, the rate of change changes faster over time. If it's down and the function is decrescent in that interval, 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 it.

Careful! In the same way [math]\displaystyle{ f'(x) = 0 }[/math] is a necessary but insufficient condition for a point to be a local maximum or minimum. [math]\displaystyle{ f''(x) = 0 }[/math] is a necessary but also insufficient condition for a point to be an inflection point.

If [math]\displaystyle{ f }[/math] is twice differentiable 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.

(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 if the concavity is downwards. Else, the concavity is upwards and the function is above its linear approximation. I'm going to follow Stewart's textbook and prove the first half.

Let [math]\displaystyle{ a }[/math] be a number that is in the interval [math]\displaystyle{ I }[/math]. We must show that [math]\displaystyle{ f(x) \gt T(x) }[/math], where [math]\displaystyle{ T(x) = f(a) - f'(a)(x - a) }[/math] for all [math]\displaystyle{ x \in I }[/math], except for the intersection where [math]\displaystyle{ f(x) = T(x) }[/math].

If [math]\displaystyle{ x \gt a }[/math], applying the Mean Value Theorem to [math]\displaystyle{ f }[/math] on the interval [math]\displaystyle{ [a, \ x] }[/math], we get a number [math]\displaystyle{ c }[/math], with [math]\displaystyle{ a \lt c \lt x }[/math], such that

[math]\displaystyle{ f(x) - f(a) = f'(c)(x - a) }[/math]

Since [math]\displaystyle{ f'' \gt 0 }[/math] on [math]\displaystyle{ I }[/math] we know that [math]\displaystyle{ f' }[/math] is crescent on [math]\displaystyle{ I }[/math]. Thus, since [math]\displaystyle{ a \lt c }[/math], we have

[math]\displaystyle{ f'(a) \lt f'(c) }[/math]

(to understand why this is true, look at the graph and pay attention to the fact that [math]\displaystyle{ c }[/math] is somewhere after [math]\displaystyle{ a }[/math] and before [math]\displaystyle{ x }[/math])

We know that [math]\displaystyle{ x - a \gt 0 }[/math], therefore if we multiply both sides by a positive number the inequality isn't changed

[math]\displaystyle{ (x - a)f'(a) \lt f'(c)(x -a) }[/math]

Adding a number to both sides should keep the inequality true

[math]\displaystyle{ f(a) + (x - a)f'(a) \lt f'(c)(x -a) + f(a) }[/math]

Looking back, we can substitute the right side of the equation with [math]\displaystyle{ f(x) }[/math], which results in

[math]\displaystyle{ f(x) \gt f(a) + (x - a)f'(a) }[/math]

the result that we wanted to prove. With a similar reasoning we repeat the process for the case of [math]\displaystyle{ x \lt a }[/math]. Careful that the next time we have that [math]\displaystyle{ x - a \lt 0 }[/math].

I have a textbook that does the proof in a slightly different manner. It considers the difference between the function and its linear approximation. If the linear approximation is below the function, then we have to prove that the difference is positive. Else, with the linear approximation being above the function, we have to prove that the difference is negative.

Note: by analogy one may think we could use the third derivative to know whether a point where [math]\displaystyle{ f''(x) = 0 }[/math] is an inflection point or not, by comparing to what we do with local maximums and minimums. This is not the case because an inflection point can never coincide with a maximum or a minimum. By definition we have a point where the concavity changes from upwards to downwards or vice-versa. In regular Euclidean geometry it's impossible for this point to be a maximum or a minimum (I have no idea if there exists a non-Euclidean geometry where this is possible).

The information that we could extract from the third derivative would be finding the point where the derivative changes from crescent to decrescent or vice-versa, which may or may not coincide with a local maximum or minimum of the derivative itself.