Problem : 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 : Here, given function is addition of two trigonometric functions. As we know least and greatest values of sine and cosine functions, we shall attempt to reduce given function in terms of either sine or cosine function (note that the algorithm for reducing addition of sine and cosine functions as presented here is a standard algorithm. We should also note that this algorithm, as a matter of fact, is 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 : 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 range of the function.

Solution : “A” is an interval. We are required to find range of function. In order to find range, however, we would need to determine least and greatest function values. Towards this, we need to know the nature of function in the interval. For this, we find derivative of function and determine 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 correspond to least and greatest values of the function and as such represent the range of the function. Now,

⇒ 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,

⇒ Range = f A = [ 1 2 − π 3 − π 2 9 , 3 2 − π 6 − π 2 36 ] ⇒ Range = f A = [ 1 2 − π 3 − π 2 9 , 3 2 − π 6 − π 2 36 ]

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,∞), find the range of the function.

Solution :

Statement of the problem : The function contains greatest integer function in its denominator, whereas numerator of functon is an exponential function. Greatest integer function, GIF, is defined in discontinuous intervals of 1. It returns integral value equal to the starting value of discontinuous interval. The function, however, is continuous in sub-intervals of 1 i.e. in a finite interval. It is clear that function values and function nature will change in accordance with the values of GIF in sub-intervals of the domain. We need to determine nature of function in few of the initial sub-intervals and extrapolate the results to determine the range of function as required.

In order to determine nature of function, we write function and its derivative in the initial sub-intervals of the given domain. Note that domain of function starts with 0.

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

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The derivatives in each interval are positive for values of “x” in the domain. It means that function is strictly increasing in each of the intervals. There is no minimum and maximum as monotonic nature of function does not change. The least and greatest values, therefore, correspond to end function values in each finite sub-intervals. Using these least and greatest values, we determine range of initial sub-intervals as given here :

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

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The plot of the function is drawn in the figure. Note that function values in adjacent intervals overlap each other, but exceeds the greatest value in the preceding intervals. Clearly, function value begins from “1” and continues towards infinity. Hence, range of the function is :

Figure 1: The range is divided in multiple overlapping intervals.

Range of the function

[ 1, ∞ ) [ 1, ∞ )

Problem : Find range of continuous function :

f x = x − 1 + 5 − x f x = x − 1 + 5 − x

Solution : Given function is continuous. Also, it involves square root of polynomial. It means that function is defined in sub interval of real number set R. We, therefore, need to find the domain of the function first. Then we investigate the nature of the function in its domain and determine least (a) and greatest (b) function values.

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 intersection of individual domains and is a finite interval [1,5]. Now, we differentiate the function with respect to independent variable to determine the nature of function :

⇒ 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 test value, x = 4, f’(x) is negative. Hence, corresponding sign scheme of f’(x) is as given here :

Figure 2: The value at the point of change in the nature of function corresponds to greatest value of the function.

Sign scheme of derivative expression

Since function is increasing to the left and decreasing to the right of root value, the function value at root value is maximum. Clearly, there is only one maximum. Thus, it is greatest value also. Further, either of two function values at end points is least value of the function in the domain . Now, function values at end points and root point 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 is greatest value and either values at x = 1 and 5 is least. Hence, range of the function is :

[ 2, 2 2 ] [ 2, 2 2 ]

Problem : 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 outermost square bracket, here, represents greatest integer function. Find range of the function.

Solution : The argument of arcsine function is a square root. But we know that domain of arcsine function is [-1,1]. If we know the least or greatest value of quadratic expression within sqaure root, then we shall be able to futher narrow down the value of argument of arcsine function. The derivative of expression within square root is :

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 test poin, x = 1 is 3, which is a positive number. Thus, corresponding sign scheme of the derivative expression is shown in the figure.

Figure 3: The value at the point of change in the nature of function corresponds to the least value of the function.

Sign scheme of derivative expression

The polynomial function is strictly decreasing to the left and strictly increasing to the right of root value. This means that function value at root point is minimum. Also, as there is only one minimum, this value is the least function value in the 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 argument of arcsine function is “1”. Also, square root is a positive number. 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 }