(a) A function $f$ is said to have a maximum value in I, if there exists a point a in I such that $f(a) \geq f (x ) \; \forall x \in I$.
The number f (a) is called the maximum value of function $f$ in I and the point a is called a point of maximum value of f in I.
(b) A function f is said to have a minimum value in I, if there exists a point a in I such that $f (a) \leq f (x) \; \forall x \in I$.
The number f (a), in this case, is called the minimum value of f in I and the point a, in this case, is called a point of minimum value of f in I.
Example 1
Find the maxima and minima of Below function
$y=x^2$ for $x \in R$ Solution
Here is the graph of the function
we have f(x) = 0 if x = 0. Also $f (x) \geq 0$, for all $x \in R$.
Therefore, the minimum value of f is 0 and the point of minimum value of f is x = 0. Further, it may be observed from the graph of the function that f has no maximum value and hence no point of maximum value of f in R
Extreme Value of the Function
A function f is said to have an extreme value in I if there exists a point a in I such that f (a) is either a maximum value or a minimum value of f in I.
The number f (a), in this case, is called an extreme value of f in I and the point a is called an extreme point.
Local Maxima and Minima of functions
In last section we talked about the maximum and minimun values of the function,But there might be other points in the domain of the Function where it does not becomes maximum and minimum but it becomes maximum and minumum with respect to the neigbhouring point.
These point are call Local Minima and Local Maximum
Let f be a real valued function and let c be an interior point in the domain of f. Then
(a) c is called a point of local maxima if there is an h > 0 such that $f (c) > f (x), \; \forall x \in (c - h, c + h), x \ne c $
The value f (c) is called the local maximum value of f.
(b) c is called a point of local minima if there is an h > 0 such that $f (c) < f (x), \; \forall x \in (c - h, c + h), x \ne c$
The value f (c) is called the local minimum value of f .
A function might have many local maxima and minima
Theorem on Local Maxima and Minima of Functions
Let f be a function defined on an open interval I. Suppose $c \in I$ be any point. If f has a local maxima or a local minima at x = c, then either $f'(c) = 0$ or f is not differentiable at c
Critical point of Function
A point c in the domain of a function f at which either $f'(c) = 0$ or f is not differentiable is called a critical point of f. Note that if f is continuous at c and $f'(c) = 0$, then there exists
an h > 0 such that f is differentiable in the interval (c - h, c + h).
First Derivative Test for Local Maxima and Minima
Let f be a function defined on an open interval I. Let f be continuous at a critical point c in I. Then
(i) If $f'(x)$ changes sign from positive to negative as x increases through c, i.e., if $f'(x) > 0$ at every point sufficiently close to and to the left of c, and $f'(x) < 0$ at every point sufficiently close to and to the right of c, then c is a point of local maxima.
(ii) If $f'(x)$ changes sign from negative to positive as x increases through c, i.e., if $f'(x) < 0$ at every point sufficiently close to and to the left of c, and $f'(x) > 0$ at every point sufficiently close to and to the right of c, then c is a point of local minima.
(iii) If f ′(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima. Infact, such a point is called point of inflection
Working Run to find Local Maximum and Minimum using First Derivative test
Step 1: Find all critical points of f in the interval, i.e., find points x where either f'(x) =0 or f is not differentiable. Step 2: For all the critical point obtained above .
if \( f'(x) \) changes from positive to negative as x increase from left to right of that point,then that point is Local Maxima
if \( f'(x) \) changes from negative to positive as x increase from left to right to that point , then that point is Local Minima
Second Derivative Test for Local Maxima and Minima
Let f be a function defined on an interval I and $c \in I$. Let f be twice differentiable at c. Then
(i) x = c is a point of local maxima if $f'(c) = 0$ and $f''(c) < 0$ The value f (c) is local maximum value of function $f$ .
(ii) x = c is a point of local minima if $f'(c) = 0$ and $f''(c) > 0$ In this case, f (c) is local minimum value of function $f$ .
(iii) The test fails if $f'(c) = 0$ and $f''(c) = 0$. In this case, we go back to the first derivative test and find whether c is a point of local maxima, local minima or a point of inflexion
Working Run to find Local Maximum and Minimum using both First and Second Derivative
Step 1: Find all critical points of f in the interval, i.e., find points x where either f'(x) =0 or f is not differentiable. Step 2: Find the second derivative test Step 3: Calculate the second derivative for all the critical points
- If \( f''(x) < 0 \) at a point where \( f'(x) = 0 \), \( f(x) \) has a local maximum at that point.
- If \( f''(x) > 0 \) at a point where \( f'(x) = 0 \), \( f(x) \) has a local minimum at that point. Step 4: If f'(x) =0 and \( f''(x) = 0 \) ,then we can use either of these technique
(i) then we go back to the first derivative test and find whether c is a point of local maxima, local minima or a point of inflexion
- Local Maxima: \( f'(x) \) changes from positive to negative.
- Local Minima: \( f'(x) \) changes from negative to positive.
(ii) Find the third derivative and if \( f'''(a) \ne 0 \), then this point is point of inflexion
If \( f'''(a) = 0 \),then find fourth derivative,if \( f''''(a) < 0 \) then point a is local maximum and if \( f''''(a) > 0 \) then point a is local minimum
Maximum and Minimum Values of a Function in a Closed Interval
First lets check few concepts here Absolute Maximum
This the maximum value of the function in the closed interval
This is also called global maximum or greatest value
Absolute Minimum
This the minimum value of the function in the closed interval
This is also called global minimum or least value.
Now lets see the function graph in closed interver [a,d]
Here f(a) : Absolute Minimum in the closed Interval
f(b) : Local Maxima
f(c): Local Minima
f(d): Absolute Maximum
Working Rule to find Maximum and Minimum Values of a Function in a Closed Interval
Step 1: Find all critical points of f in the interval, i.e., find points x where either f'(x) =0 or f is not differentiable. Step 2: Take the end points of the interval. Step 3: At all these points (listed in Step 1 and 2), calculate the values of f . Step 4: Identify the maximum and minimum values of f out of the values calculated in Step 3. This maximum value will be the absolute maximum (greatest) value of f and the minimum value will be the absolute minimum (least) value of f
Solved Examples
Question 1
Find the maxima and minima of the function \(f(x) = x^3 - 3x^2 - 9x + 5\). Solution
Step 1: Find critical points
\[f'(x) = 3x^2 - 6x - 9\]
Set \(f'(x) = 0\)\
\[3x^2 - 6x - 9 = 0\]
\[x^2 - 2x - 3 = 0\]
\[x = 3, x = -1\]
Step 2: Use the second derivative test.
\[f''(x) = 6x - 6\]
For \(x = 3\), \(f''(3) = 12 > 0\). This means \(x = 3\) is a point of minima.
For \(x = -1\), \(f''(-1) = -12 < 0\). This indicates \(x = -1\) is a point of maxima.
Question 2
Determine the maxima and minima of the function \(f(x) = x^2 e^{-x}\) for \(x \geq 0\). Solution
Step 1: Find critical points
\[f'(x) = (2x - x^2) e^{-x}\]
Set \(f'(x) = 0\).
\[2x - x^2 = 0\]
\[x = 0, x = 2\]
Step 2: Second derivative test.
\[f''(x) = (x^2 - 4x + 2) e^{-x}\]
For \(x = 0\), \(f''(0) = 2 > 0\). Minima.
For \(x = 2\), \(f''(2) = -4e^{-2} < 0\). Maxima.
Therefore
The function has a maximum at \(x = 2\) and a minimum at \(x = 0\).
