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Derivative & Turning Point

confusedatmath

New member
Jan 2, 2014
14
Can someone explain the link between the turning point (local max, min & stationary point of inflection) and it's relationship to derivatives.

Let me clarify what I understand (feel free to correct me).

If we derive an equation and let it = 0, the value of x is some kind of turning point.

To find out what kind, we derive again, and sub that value of x , and look for the following.

if x > 0 it is a local min

if x < 0 is it a local max

if x = 0 it is an inflection point.

example.

y= (x-1)^3 + 1

is this a local max/min/inflection point at (1,1)

so

derive 1st time = 3(x-1)^2 = 3x^2 -6x +3
we let it equal 0 so

0 = 3x^2 -6x +3
x =1

then we derive again to know what kind of turning point.

derive 2nd time = 6x -6

sub x=1
6(1)-6 = 0

therefore it is an inflection point??

is this right... hmmmm
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Since this thread discusses applications of the derivative, I have moved it here to our Calculus subforum. Unfortunately, I have to go now, but if no one has addressed your questions by the time I return, I will do so then. :D
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
If we derive an equation and let it = 0, the value of x is some kind of turning point.
I would use the terminology "differentiate" or "take the derivative of." Deriving refers to the act of algebraically finding an expression.

Now, the fact that the derivative of a function is equal to zero for some value of the independent variable is a necessary, but not sufficient condition. The derivative must change its sign across this critical value.

For example, consider the function:

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

Differentiating with respect to $x$, we find:

\(\displaystyle f'(x)=3x^2\)

Now, this derivative has a root at $x=0$, however, this roots is of even multiplicity, and so we know the derivative does not change sign across this critical value, and so there is not a turning point there. The derivative is positive on either side of this critical value, so we know the function is increasing on either side, and so does not turn or change direction.

To find out what kind, we derive again, and sub that value of x , and look for the following.

if x > 0 it is a local min

if x < 0 is it a local max

if x = 0 it is an inflection point.
I think you mean to use $f''(x)$ here rather than $x$. It is the value of the second derivative which we analyze at the root(s) of the first derivative.

Also, if the value of the second derivative is zero at one of the roots of the first derivative, then all we can say is that no conclusion may be drawn from the second derivative test for relative extrema.

example.

y= (x-1)^3 + 1

is this a local max/min/inflection point at (1,1)

so

derive 1st time = 3(x-1)^2 = 3x^2 -6x +3
we let it equal 0 so

0 = 3x^2 -6x +3
x =1

then we derive again to know what kind of turning point.

derive 2nd time = 6x -6

sub x=1
6(1)-6 = 0

therefore it is an inflection point??

is this right... hmmmm
In the case of the function you cite, yes that is an inflection point, but only because the sign of the second derivative changed across the root. We know the sign will change because the root is of odd multiplicity, the multiplicity being 1 in this case. The second derivative is linear, so its sign must change across its root.

If we consider the quartic function:

\(\displaystyle f(x)=x^4\)

then we find:

\(\displaystyle f''(x)=12x^2\)

Now, this second derivative has a root at $x=0$, however it is a root of multiplicity 2, so we know the sign of the second derivative does not change across this root. The second derivative is positive on either side of the root, so we know the function is concave up in both intervals, and the point (0,0) is not a point of inflection.
 

confusedatmath

New member
Jan 2, 2014
14
There are a couple of things i didn't get in your explanation, and its due to my lack of vocab in calculus. :p

this derivative has a root at x=0 (what do you mean root?)
this roots is of even multiplicity (again, what is root/ multiplicity?)
from the second derivative test for relative extrema (relative extrema?)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
There are a couple of things i didn't get in your explanation, and its due to my lack of vocab in calculus. :p

this derivative has a root at x=0 (what do you mean root?)
this roots is of even multiplicity (again, what is root/ multiplicity?)
from the second derivative test for relative extrema (relative extrema?)
A root or zero of a function is a value of the independent variable (traditionally $x$) for which the value of the function is zero.

The number of instances of a particular root is its multiplicity. For example, the function:

\(\displaystyle f(x)=x(x-1)^2(x-2)^5(x-3)^n\)

Has the following roots:

\(\displaystyle x=0\) of multiplicity 1

\(\displaystyle x=1\) of multiplicity 2

\(\displaystyle x=2\) of multiplicity 5

\(\displaystyle x=3\) of multiplicity $n$

Relative extrema are either maxima or minima, also known as turning points. They are places where the derivative of a function is zero, and the sign of the first derivative is different in the intervals on either side.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
A root or zero of a function is a value of the independent variable (traditionally $x$) for which the value of the function is zero.
If I were very picky (which I am, of course), I would disagree with this. Yes, a "zero" of a function is a value of the variable that makes the function equal to 0. But the word "root" does not apply to a function, it applies to an equation. A zero of function f is a value of x such that f(x)= 0. And so a zero of f is root of the equation f(x)= 0.

The number of instances of a particular root is its multiplicity. For example, the function:

\(\displaystyle f(x)=x(x-1)^2(x-2)^5(x-3)^n\)

Has the following roots:

\(\displaystyle x=0\) of multiplicity 1

\(\displaystyle x=1\) of multiplicity 2

\(\displaystyle x=2\) of multiplicity 5

\(\displaystyle x=3\) of multiplicity $n$

Relative extrema are either maxima or minima, also known as turning points. They are places where the derivative of a function is zero, and the sign of the first derivative is different in the intervals on either side.