First Derivative Test / Intervals of Increase, Decrease / Concavity test

[StintedVisions]

So in attempts to ensure I pass the class I've been reading through the sections of the chapter we have a test on this Friday and I believe I understand everything correctly but had a question that the book (I think) hints at, but doesn't say blatantly.

The Increasing/decreasing test states that for a function that meets all of the necessary criteria:
If f' is negative for a test point chosen in the interval being tested the slope is decreasing
If f' is positive for a test point chosen in the interval being tested the slope is increasing

So the First Derivative test states that for a function that meets all of the necessary criteria:
If f' changes sign from + to - the graph has a local max
If f' changes sign from - to + the graph has a local min
If f' doesn't change sign the test is inconclusive

And you can tell concavity by observing that:
If f' is increasing then f is concave up
If f' is decreasing then f is concave down

I just want to make sure I have this correct, that these are all basically the same test but telling you three different pieces of information? I believe I'm understanding it correctly but please correct me if I'm wrong.

I'm also not understanding the purpose of the concavity test.
if f" > 0 on I then f is concave up
if f" < 0 on I then f is concave down

If you can already tell this from the information from the first derivative test what's the point of testing for concavity?

 

[MarkFL]

The first derivative does not tell you about concavity, but the second derivative does. Also, at points where the first derivative is zero, the sign of the second derivative at that point can tell you about the nature of the possible extremum there.

The first derivative can tell you whether a function is increasing or decreasing, but it is the second derivative that tells you about the rate of this change, which is concavity.

 

[HallsofIvy]

The simplest way to verify all of that is to look at a simple function: \(\displaystyle y= x^2\). That obviously has a minimum at (0, 0) because its graph is a parabola opening upward with vertex at (0, 0). Its first derivative is \(\displaystyle y'= 2x\). That is, of course, 0 at x= 0. For x< 0, y' is negative, for x> 0, y'> 0. That is y' is changing from negative to positive. Geometrically, y' negative means y is decreasing, y' positive means y is increasing. So y is decreasing as x approaches 0 from below then starts increasing as x become positive. Clearly that means that y has a minimum at x= 0.

And, of course, because y' goes from negative to positive, y' is increasing and so its derivative, y'', is positive.

 

[lookagain]

> > > The first derivative does not tell you about concavity, < < <


but the second derivative does. Also, at points where the first derivative is zero,


the sign of the second derivative at that point can tell you about the nature of the possible extremum there.


The first derivative can tell you whether a function is increasing or decreasing,


but it is the second derivative that tells you about the rate of this change, which is concavity.

.


But consider the graphs of these functions in the real xy-plane for examples.
(Let us see if the conclusions follow from the information above each conclusion.)
These examples do not use the second derivative to determine their respective concavities:


\(\displaystyle y \ = \ \sqrt[3]{x^2 \ } \)

It is continuous everwhere. The first derivative is undefined at x = 0,
but there is a vertical tangent there. The first derivative is negative for x < 0,
and positive for x > 0.

Conclusion: It is concave down for x < 0 and for x > 0.


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\(\displaystyle y \ = \ \sqrt[3]{x \ } \)

It is continuous everywhere. The first derivative is undefined at x = 0,
but there is a vertical tangent there. The first derivative is positive for
both x < 0 and for x > 0.

Conclusion: It is concave up for x < 0 and concave down for x > 0.



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\(\displaystyle y \ = \ x^3 \)

It is continuous everywhere. The first derivative is zero at x = 0.
The first derivative is positive for both x < 0 and for x > 0.


Conclusion: It is concave down for x < 0 and concave up for x > 0.


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\(\displaystyle y \ = \ \dfrac{1}{x} \)

The first derivative is undefined at x = 0. There is no point at x = 0, because the
function is undefined there, but elsewhere it is defined. The first derivative is
negative for both x < 0 and for x > 0. There is a vertical asymptote at x = 0
and a horizontal asmptote at y = 0.

Conclusion: It is concave down for x < 0 and concave up for x > 0.


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\(\displaystyle y \ = \ \dfrac{1}{x^2} \)

The first derivative is undefined at x = 0. There is no point at x = 0,
because the function is undefined there, but elsewhere it is defined.
The first derivative is positive for x < 0 and negative for x > 0.
There is a vertical asymptote at x = 0 and a horizontal asymptote
at y = 0.

Conclusion: It is concave up for both x < 0 and for x > 0.


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\(\displaystyle y \ = \ \sqrt{x \ } \)

There are no points for x < 0, but elsewhere there are. The first derivative is
undefined at x = 0. The first derivative is positive for x > 0. As there is a point
at x = 0 and the first derivative there is undefined, there is a vertical tangent
at x = 0.

Conclusion: It is concave down for x > 0.


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\(\displaystyle y \ = \ e^x \)

It is continuous everwhere. The first derivative is positive everywhere. The function
is everywhere positive, and it has a horizontal asymptote at y = 0.

Conclusion: It is concave up everywhere.


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\(\displaystyle y \ = \ \ln{(x)} \ \ \ \ \ \ (the \ \ natural \ \ log \ \ of \ \ x) \)

It is continuous/exists for x > 0. As the first derivative is 1/x and the domain
of the original function is x > 0, then the first derivative is positive for x > 0. As the
first derivative is undefined at x = 0 and the domain is x > 0, there is a vertical
asymptote at x = 0.

Conclusion: It is concave down for x > 0.