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Least and greatest function values

Module by: Sunil Kumar Singh

Least and greatest values of a function represents range of a function in a finite interval. This fact can be used as a very useful technique to determine range of a function. We realize that it is always not possible to solve a function for independent variable “x” to find range. The problem is more pronounced, if the function is transcendental (trigonometric, exponential, logarithmic, inverse trigonometric etc.). In such cases, we can employ the monotonous property of a function to find range, provided function is continuous in a given interval.
The end values of a function, which continuously increases or decreases in a finite domain, represent minimum (least) and maximum (greatest) values of the function. The lower value represents the least value and higher value represents the greatest value. These least and greatest values correspond to end values of interval representing range of the function.
A continuous function, however, may change nature in a given domain interval. If function has specific nature (increasing of decreasing) in sub-intervals of the domain, then we can determine least and greatest values of the function corresponding to points where it changes nature or where the function is bounded in the interval.
The derivative of function is helpful in this regard. Using sign scheme, we can find different intervals in which derivative is either positive or negative. The function is an increasing function in positive sub-interval, whereas it is decreasing in negative interval. The point about which function changes its nature represents either least or greatest value of the function.
In order to understand the requirement of the method, let us consider the plot of “ x 2 x 2 ” as shown here. The function has least value at x = 0. Where does it attain the greatest value? We can see that the plot extends to a very large value “y” on either side of y-axis.
Square function
r1.gif
Figure 1: The plot extends to a very large value “y” on either side of y-axis.
Clearly, we need to consider a finite domain of the function such that we can assess the greatest value(s) of the function, which is finite. It should be clearly understood that this technique suits evaluation of range of functions, which is continuous in a finite domain or is continuous in finite intervals of the domain.
Working rules
  • If the function is continuous in a finite interval, then find the derivative of the given function.
  • Determine least (a) and greatest (b) function values with the help of sign scheme of the corresponding derivative expression (or otherwise). We should note that we can determine least and greatest values of the function even without resorting to differentiation.
  • The range of the function is interval between “a” and “b”.
Note: The problems in this module involves finding range of function, which is based on the concept of increasing or decreasing function. This requires use of derivative of the given function. As derivative is not yet covered, this module can be skipped by readers not exposed to the concept of derivatives and revisit the module after studying the same.

Problem 1: Find range of continuous function :
f x = x − 1 + 5 − x f x = x − 1 + 5 − x
Solution : First we need to find the domain of the function and see whether it is finite or not? Then we determine least (a) and greatest (b) function values with the help of sign scheme (or otherwise) to determine range of the function.
For individual square roots to be real,
x - 1 ≥ 0 ⇒ x ≥ = 1 x - 1 ≥ 0 ⇒ x ≥ = 1
5 - x ≥ 0 ⇒ x ≤ 5 5 - x ≥ 0 ⇒ x ≤ 5
Hence, domain of the function is a finite interval [1,5]. Now, we differentiate the function with respect to independent variable as :
⇒ f ′ x = 1 2 x − 1 − 1 2 5 − x ⇒ f ′ x = 1 2 x − 1 − 1 2 5 − x
In order to draw sign scheme, we find the roots of the equation by putting f’(x) = 0 and solving for “x” as :
⇒ 2 x − 1 = 2 5 − x ⇒ 2 x − 1 = 2 5 − x
⇒ x − 1 = 5 − x ⇒ 2 x = 6 ⇒ x = 3 ⇒ x − 1 = 5 − x ⇒ 2 x = 6 ⇒ x = 3
For x = 4, f’(x) is negative. Corresponding sign scheme of f’(x) is as given here :
Sign scheme of derivative expression
r2.gif
Figure 2: The value at the point of change in the nature of function corresponds to greatest value of the function.
The function values are :
⇒ f 1 = 0 + 2 = 2 ⇒ f 1 = 0 + 2 = 2
⇒ f 3 = 2 + 2 = 2 2 ⇒ f 3 = 2 + 2 = 2 2
⇒ f 5 = 2 + 0 = 2 ⇒ f 5 = 2 + 0 = 2
Clearly, function at x = 3 attains greatest value and least values at x = 1 and 5. Hence, range of the function is :
[ 2, 2 2 ] [ 2, 2 2 ]

Problem 2: Let A = [ π 6 , π 3 ] A = [ π 6 , π 3 ] . If a continuous function is defined as :
f x = cos x − x 1 + x f x = cos x − x 1 + x
Find f(A).
Solution :
Statement of the problem : “A” is an interval. We are required to find f(A). It means that we have to find values of function corresponding to interval specified by “A”. The interval of values denoted by “f(A)” obviously is “range” of the function for domain specified by “A”.
In order to find range, we would need to find derivative and its sign. This will enable us to know whether the function is increasing or decreasing. Now,
⇒ f ′ x = − sin x − 1 − 2 x ⇒ f ′ x = − sin x − 1 − 2 x
The terms “sinx” and “2x” are positive for values of “x” in the interval specified by “A”.
Therefore, we can conclude that the derivative is negative. It means that the function is a decreasing function in the entire interval. The end values of the function corresponds to least and greatest values of the function :
⇒ f x max = f π / 6 = cos π 6 − π 6 − π 6 2 = 3 2 − π 6 − π 2 36 ⇒ f x max = f π / 6 = cos π 6 − π 6 − π 6 2 = 3 2 − π 6 − π 2 36
⇒ f x min = f π / 3 = cos π 3 − π 3 − π 3 2 = 1 2 − π 3 − π 2 9 ⇒ f x min = f π / 3 = cos π 3 − π 3 − π 3 2 = 1 2 − π 3 − π 2 9
Hence,
⇒ f A = [ 1 2 − π 3 − π 2 9 , 3 2 − π 6 − π 2 36 ] ⇒ f A = [ 1 2 − π 3 − π 2 9 , 3 2 − π 6 − π 2 36 ]

Problem 4: Find the range of the continuous function given by :
f x = a cos x + b sin x f x = a cos x + b sin x
where “a” and “b” are constants.
Solution :
Statement of the problem : The given function is addition of two trigonometric functions. We have to find the range of the given function.
We observe that the range of sine and cosine functions are bounded between “-1” and “1”. Hence, we should expect a range for the given function which is also bounded between minimum and maximum values.
In order to estimate least and greatest values, we reduce the given function in terms of one of two trigonometric functions (note that the algorithm for reducing addition of sine and cosine functions as presented is a standard algorithm. We may find that this algorithm, as a matter of fact, is extensively used in analyzing superposition principle of waves) :
Let a = r cos α and b = r sin α Let a = r cos α and b = r sin α
where,
r = a 2 + b 2 r = a 2 + b 2
Substituting in the given function, we have :
f x = r cos α cos x + r sin α sin x = r cos x − α = a 2 + b 2 cos x − α f x = r cos α cos x + r sin α sin x = r cos x − α = a 2 + b 2 cos x − α
We know that minimum and maximum values of cosine function are "-1" and "1" respectively. Hence,
f x min = - a 2 + b 2 f x min = - a 2 + b 2
f x max = a 2 + b 2 f x max = a 2 + b 2
Therefore, range of the given function is :
Range = [ - a 2 + b 2 , a 2 + b 2 ] Range = [ - a 2 + b 2 , a 2 + b 2 ]

Problem 4: A function is defined as :
f x = [ log e { sin - 1 x 2 + x + 1 } ] f x = [ log e { sin - 1 x 2 + x + 1 } ]
The square bracket, here, represents greatest integer function. If domain of the function is [-1,0], find its range.
Solution :
Statement of the problem : The function is greatest integer function. Its domain is a finite interval. Since greatest integer function returns greatest integer, it is possible that function return a single value as expression. In the nutshell, we need to estimate the value of function for values of “x” in this interval.
Now, the argument of arcsine function is a square root. The derivative of expression within square root is :
f ′ x = 2 x + 1 f ′ x = 2 x + 1
The root of this expression when equated to zero is x = -1/2. The function value at x = 1 is 3, which is a positive number. Corresponding sign scheme of the derivative expression is shown in the figure.
Sign scheme of derivative expression
r4.gif
Figure 3: The value at the point of change in the nature of function corresponds to the least value of the function.
Clearly, the value at x=-1/2 corresponds to least value of the expression for the values of “x” in the given domain. Thus, least value of square root is :
⇒ x 2 + x + 1 = { − 1 2 2 + − 1 2 + 1 } = 3 2 ⇒ x 2 + x + 1 = { − 1 2 2 + − 1 2 + 1 } = 3 2
But, we know that the maximum value of the arcsine function is “1”. It means that :
x 2 + x + 1 ≤ 1 x 2 + x + 1 ≤ 1
Thus, we conclude that the value of square root expression should lie in the interval given by :
⇒ 3 2 ≤ x 2 + x + 1 ≤ 1 ⇒ 3 2 ≤ x 2 + x + 1 ≤ 1
We see that arcsine is an increasing function within the interval specified by domain of the function. Hence, we can take sine inverse of each term without any change in the direction of inequality :
⇒ sin - 1 3 2 ≤ sin - 1 x 2 + x + 1 ≤ sin - 1 1 ⇒ sin - 1 3 2 ≤ sin - 1 x 2 + x + 1 ≤ sin - 1 1
⇒ π 3 ≤ sin - 1 x 2 + x + 1 ≤ π 2 ⇒ π 3 ≤ sin - 1 x 2 + x + 1 ≤ π 2
Logarithmic function is also an increasing function. Hence, we can take logarithm of each term without any change in the direction of inequality :
log e π 3 ≤ log e sin - 1 x 2 + x + 1 ≤ log e π / 2 log e π 3 ≤ log e sin - 1 x 2 + x + 1 ≤ log e π / 2
Since log e π / 2 < 1 log e π / 2 < 1 and log e π / 3 > 0 log e π / 3 > 0 , we can state the inequality as :
0 < log e sin - 1 x 2 + x + 1 < 1 0 < log e sin - 1 x 2 + x + 1 < 1
We know that greatest integer function returns "0" in the interval (0,1). Hence,
f x = [ log e sin - 1 x 2 + x + 1 ] = 0 f x = [ log e sin - 1 x 2 + x + 1 ] = 0
Clearly, function returns zero for all values of “x” in the domain of the function. Therefore, range of the function is :
Range = { 0 } Range = { 0 }

Problem 5: A function is defined as :
f x = e x 1 + [ x ] f x = e x 1 + [ x ]
If domain of the function is [0,π), find the range of the function.
Solution :
Statement of the problem : The domain of the function is given. The domain of the function contains greatest integer function, which changes values in different intervals of “x”. Clearly, the function is continuous in sub-intervals. We need to estimate range of the function in each of the sub-intervals.
Within given domain, we write function and its derivative in different intervals (corresponding to intervals of greatest integer function) as :
0 ≤ x < 1, f x = e x , f ′ x = e x 0 ≤ x < 1, f x = e x , f ′ x = e x
1 ≤ x < 2, f x = e x 2 , f ′ x = e x 2 1 ≤ x < 2, f x = e x 2 , f ′ x = e x 2
2 ≤ x < 3, f x = e x 3 , f ′ x = e x 3 2 ≤ x < 3, f x = e x 3 , f ′ x = e x 3
------------------------------------- -------------------------------------
The derivatives in each interval are positive for values of “x”. It means that function is increasing in each of the intervals. Thus, we determine the corresponding intervals of function values by determining minimum and maximum values in each of these intervals.
0 ≤ x < 1 ⇒ e 0 ≤ y < e 1 ⇒ 1 ≤ y < e 0 ≤ x < 1 ⇒ e 0 ≤ y < e 1 ⇒ 1 ≤ y < e
1 ≤ x < 2 ⇒ e 2 ≤ y < e 2 2 1 ≤ x < 2 ⇒ e 2 ≤ y < e 2 2
2 ≤ x < 3 ⇒ e 2 3 ≤ y < e 3 3 2 ≤ x < 3 ⇒ e 2 3 ≤ y < e 3 3
------------------------------------- -------------------------------------
The plot of the function is drawn in the figure. Clearly, range begins from “1” and continues towards infinity. Hence, range of the function is :
Range of the function
r5.gif
Figure 4: The range is divided in multiple overlapping intervals.
[ 1, ∞ ) [ 1, ∞ )

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