Maximum And Minimum Values | Critical Points | Concavity | Inflection Point

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We can find maximum and minimum values of a function f(x) at its critical points. Critical points of a function are those where its derivative either zero or undefined (i. e. corner of a function). So the critical points occur where ever the slope of the function is completely flat. From critical point we can define the maximum and minimum values of a function, but it is to remember that not in all critical point there exists maximum or minimum. learn more. In this article we will also discuss about concavity of function, and its inflection point.

There is concept of global maximum and global minimum, also called absolute maximum, and absolute minimum. Global maximum is the point which has a maximum value for the entire domain of the function. Similarly global minimum is the point which has minimum value for the entire domain of the function. If c is a point in the domain D of a function f. Then f(c) is the
absolute maximum value of f on D if f (c) ≥ f (x) for all x in D.
absolute minimum value of f on D if f(c) ≤ f(x) for all x in D.

On the other hand Local maximum is the point which has a maximum value for a specific interval, or specific region of the function. Similarly Local minimum also has the minimum value over an specific interval of the function. The point f(c) is a
local maximum value of f if f(c) ≥ f(x) when x is near c.
local minimum value of f if f(c) ≤ f(x) when x is near c.

Now let’s find out the maximum and minimum values of a given function.

Maximum And Minimum Values| Critical Points | Concavity | Inflection Point - CS School
Figure: Maximum and Minimum

Maximum and minimum Values of function

Let a function $f(x)=3x^{4}-4x^{3}+2$. As we have learn maximum and minimum values exist in critical point of the function. For critical points slope of the function is completely flat, means derivative of the function is zero, $f'(x)=0$. Therefore, $f'(x)=12x^{3}-12x^{2}=0$. Solving for $x$ we find,$$12x^{3}+12x^{2}=0\\
=> x^{3}-x^{2}=0\\
=> x^{2}(x-1)=0\\
so, x^{2}=0, \quad\text{ } (x-1)=0\\
=> x=0, \quad\text{ } x=1$$

So, these are the critical points of the function where is the maximum and minimum. Now putting $x=0, x=1$ to our original function, we can find the maximum and minimum values of the function. But we want to know specifically at which point its maximum and minimum value exist. So, for that we need to find the second derivative of the function. If in a critical point second derivative is greater than 0, then for that point the function has the minimum value. Also if the second derivative is less than zero, then for that point the function has maximum value.

$$f″(x)=36x^{2}-24x$$

now, finding the value of second derivative at the critical point $x=0$, $f″(0)=0$. So at $x=0$, $f″(x)$ is neither positive nor negative. It might be the inflection point, means at that point concavity of the function might change direction. We are not sure about that is the inflection point. we have to test to be confirmed (we have discussed it below). Now putting $x=0$ to our main function: $f(0)=2$. So, the critical point is (0, 2). This is the maximum point (comparing with other point). Or at that critical point we have the maximum value of the function.

Again, at $x=1$ the second derivative, $f″(1)=36-24=12$, which is positive, $f″(1)>0$. Therefore at $x=1$ there exist the minimum value of the function. So the minimum value is: $f(1)=3-4+2=1$. Hence, at the critical point (1, 1) the function has the minimum value.

Concavity | Concave upward or downward

Concavity of a function means how the graph of the function bending (upward or downward). In simple words, A function is concave up when its second derivative is positive, or $f″(x)>0$. Also a function might be concave down if the second derivative is negative, $f″(x)<0$.

Maximum And Minimum Values| Critical Points | Concavity | Inflection Point
Figure: Concavity and Inflection point

Inflection Point

Inflection point is a point where the concavity of the function change direction, or sign. As we know if the second derivative of function is negative, $f″(x)<0,$ the graph is concave down. Also if the the second derivative is positive then the graph concave up. Therefore there must be a point on that the concavity change direction. As we have previously assume that at $x=0$ is the inflection point as $f″(0)=0$. THerefore let check the function concavity near around $x=0$. So, when $x=1,$ $f″(1)=36.1^{2}-24.1= 12,$ which is positive, $f″(x)>0$. Again at $x=-1,$ $f″(-1)=36(-1)^{2}-24(-1)=60$. Which means at $x=0$ the concavity doesn’t change sign. Therefore, $x=0$ is not the inflection point. So for inflection point second derivative is equal to zero.

$$f″(x)=36x^{2}-24x=0x\\
=> 12x(3x-2)=0\\
=> 12x=0 \quad\text{ | } => 3x-2=0\\
=> x=0, \quad\text{ | }=> x=\frac{2}{3}$$

As $x=0$ is not an inflection point we have tested, therefore, $x=0$ is not accepted here. So the at $x=\frac{2}{3}$ the concavity of the function change the sign. We can test this now:

Taking $x$ less than $\frac{2}{3}$. Let $x=\frac{1.9}{3}$ we get,

$$f″(\frac{1.9}{3})=-\frac{17}{25}$$

Which is negative, $f″(\frac{1.9}{3})<0$. So for this the concavity is downward. Now taking $x$ is slightly greater the $\frac{2}{3}$. Let $x=\frac{2.1}{3}$. We get,

$$f″(\frac{2.1}{3})=\frac{21}{25}$$

Which is positive $f″(\frac{1.9}{3})>0$. So, for this the concavity of the function is upward. So indeed around $x=\frac{2}{3}$ the concavity of the function change its sign. Therefore, $x=\frac{2}{3}$ is the inflection point. 

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