Problem : Find the domain of the function given by (Be aware that "x" appears as base of given logrithmic function):

f x = log x 2 f x = log x 2

Solution : By definition of logarithmic function, we know that base of logarithmic function is a positive number excluding x =1.

x > 0, x ≠ 1 x > 0, x ≠ 1

Hence, domain of the given function is :

Figure 1: Thick line represents domain of the given function.

Domain of the function

Domain = 0, ∞ − { 1 } Domain = 0, ∞ − { 1 }

or,

Domain = 0,1 ∪ { 1, ∞ } Domain = 0,1 ∪ { 1, ∞ }

Problem : Find the domain of the function given by :

f x = log 10 x 2 − 5 x + 6 x 2 + 5 x + 9 f x = log 10 x 2 − 5 x + 6 x 2 + 5 x + 9

Solution : The argument (input to the function) of logarithmic function is a rational function. We need to find values of “x” such that the argument of the function evaluates to a positive number. Hence,

⇒ x 2 − 5 x + 6 x 2 + 5 x + 9 > 0 ⇒ x 2 − 5 x + 6 x 2 + 5 x + 9 > 0

In this case, we can not apply sign scheme for the rational function as a whole. Reason is that the quadratic equation in the denominator has no real roots and as such can not be factorized in linear factors. We see that discreminant,"D", of the quadratic equation in the denominator, is negative :

⇒ D = b 2 − 4 a c = 5 2 − 4 X 1 X 9 = 25 − 36 = - 11 ⇒ D = b 2 − 4 a c = 5 2 − 4 X 1 X 9 = 25 − 36 = - 11

The quadratic expression in denominator is positive for all value of x as coefficient of squared term is positive. Clearly, sign of rational function is same as that of quadratic expression in the numerator. The coefficient of squared term of the numerator “ x 2 x 2 ”, is positive for all values of “x”. The quadratic expression in the numerator evaluates to positive for intervals beyond root values. The roots of the corresponding equal equation is :

⇒ x 2 − 2 x − 3 x + 6 = 0 ⇒ x x − 2 − 3 x − 2 = 0 ⇒ x − 2 x − 3 = 0 ⇒ x 2 − 2 x − 3 x + 6 = 0 ⇒ x x − 2 − 3 x − 2 = 0 ⇒ x − 2 x − 3 = 0

Figure 2: Thick line represents domain of the given function.

Domain of the function

x < 2 or x > 3 x < 2 or x > 3

Domain = ( - ∞ , 2 ) ∪ ( 3, ∞ ) Domain = ( - ∞ , 2 ) ∪ ( 3, ∞ )

Problem : Find the domain of the function given by :

f x = log 10 6 x − x 2 8 f x = log 10 6 x − x 2 8

Solution : The function is a square root of a logarithmic function. On the other hand argument of logarithmic function is a rational function. In order to find the domain of the given function, we first determine what values of “x” are valid for logarithmic function. Then, we apply the condition that expression within square root should be non-negative number. Domain of given function is intersection of intervals of x obtained for each of these conditions. Now, we know that argument (input to function) of logarithmic function is a positive number. This implies that we need to find the interval of “x” for which,

⇒ 6 x − x 2 8 > 0 ⇒ 6 x − x 2 8 > 0

⇒ 6 x − x 2 > 0 ⇒ 6 x − x 2 > 0

In above step, we should emphasize here that we multiply “8” and “0” and retain the inequality sign because 8>0. Now, we multiply the inequality by “-1”. Therefore, inequality sign is reversed.

⇒ x 2 − 6 x < 0 ⇒ x 2 − 6 x < 0

Here, roots of corresponding quadratic equation “ x 2 − 6 x x 2 − 6 x ” is x = 0, 6. It means that middle interval between “0 and 6” is negative as coefficient of “ x 2 x 2 ” is positive i.e. 6>0. Hence, interval satisfying the inequality is :

Figure 3: Thick line represents domain of the given function.

Domain of the function

0 < x < 6 0 < x < 6

Now, we interpret second condition according to which the whole logarithmic expression within the square root should be a non-negative number.

⇒ log 10 6 x − x 2 8 ≥ 0 ⇒ log 10 6 x − x 2 8 ≥ 0

We use the fact that if log a x ≥ y , then x ≥ a y for a > 1 if log a x ≥ y , then x ≥ a y for a > 1 . This gives us the inequality as given here,

⇒ 6 x − x 2 8 ≥ 10 0 ⇒ 6 x − x 2 8 ≥ 10 0

⇒ 6 x − x 2 8 ≥ 1 ⇒ 6 x − x 2 8 ≥ 1

⇒ 6 x − x 2 ≥ 8 ⇒ 6 x − x 2 − 8 ≥ 0 ⇒ 6 x − x 2 ≥ 8 ⇒ 6 x − x 2 − 8 ≥ 0

⇒ x 2 − 6 x + 8 ≤ 0 ⇒ x 2 − 2 x - 4 x + 8 ≤ 0 ⇒ x 2 − 6 x + 8 ≤ 0 ⇒ x 2 − 2 x - 4 x + 8 ≤ 0

⇒ x x − 2 − 4 x − 2 ≤ 0 ⇒ x − 2 x − 4 ≤ 0 ⇒ x x − 2 − 4 x − 2 ≤ 0 ⇒ x − 2 x − 4 ≤ 0

Clearly, “2” and “4” are the roots of the corresponding quadratic equation. Following sign scheme, we pick middle negative interval :

Figure 4: Thick line represents domain of the given function.

Domain of the function

2 ≤ x ≤ 4 2 ≤ x ≤ 4

Now, the interval of “x” valid for real values of “f(x)” is the one which satisfies both conditions simultaneously i.e. the interval common to two intervals determined. Hence,

Figure 5: Thick line represents domain of the given function.

Domain of the function

Domain = 0 < x < 6 ∩ 2 ≤ x ≤ 4 Domain = 0 < x < 6 ∩ 2 ≤ x ≤ 4

Domain = 2 ≤ x ≤ 4 = [ 2,4 ] Domain = 2 ≤ x ≤ 4 = [ 2,4 ]

Problem : Find the domain of the function given by :

f x = { log 0.2 x 3 + log 0.2 x 3 X log 0.2 0.0016 x + 36 } f x = { log 0.2 x 3 + log 0.2 x 3 X log 0.2 0.0016 x + 36 }

Solution : The function is square root of an expression, consisting logarithmic functions. Here, we first need to simplify expression, using logarithmic identities, before attempting to find domain of the function. Let us first simplify the middle term of the given expression, using logarithmic identities :

log 0.2 x 3 X log 0.2 0.0016 x = 3 log 0.2 x X log 0.2 0.2 4 x log 0.2 x 3 X log 0.2 0.0016 x = 3 log 0.2 x X log 0.2 0.2 4 x

⇒ log 0.2 x 3 X log 0.2 0.0016 x = 3 log 0.2 x X 4 log 0.2 0.2 + log 0.2 x ⇒ log 0.2 x 3 X log 0.2 0.0016 x = 3 log 0.2 x X 4 log 0.2 0.2 + log 0.2 x

We observe that all logarithmic functions have the base of “0.2”. Let us consider that z = log 0.2 x z = log 0.2 x , then logarithmic expression within square root is :

⇒ z 3 + 3 z 4 + z + 36 = z 3 + 3 z 2 + 12 z + 36 = z 2 z + 3 + 12 z + 3 ⇒ z 3 + 3 z 4 + z + 36 = z 3 + 3 z 2 + 12 z + 36 = z 2 z + 3 + 12 z + 3

⇒ z 3 + 3 z 4 + z + 36 = z 2 + 12 z + 3 ⇒ z 3 + 3 z 4 + z + 36 = z 2 + 12 z + 3

Now, this expression is non-negative for square root to be real. Hence,

⇒ z 2 + 12 z + 3 ≥ 0 ⇒ z 2 + 12 z + 3 ≥ 0

But, we see that z 2 + 12 z 2 + 12 is a positive number as term z 2 z 2 is positive. It means that :

⇒ z + 3 ≥ 0 ⇒ log log 0.2 x ≥ - 3 ⇒ x ≤ 0.2 - 3 ⇒ x ≤ 1 0.008 ⇒ x ≤ 125 ⇒ z + 3 ≥ 0 ⇒ log log 0.2 x ≥ - 3 ⇒ x ≤ 0.2 - 3 ⇒ x ≤ 1 0.008 ⇒ x ≤ 125

Note that we have reversed the inequality as the base is 0.2, which is less than 1. Further, we have substituted as :

z = log 0.2 x z = log 0.2 x

This logarithmic function is valid by definition for all positive value of “x”. Now, the domain of given function is the intersection of two intervals as shown in the figure.

Figure 6: Thick line represents domain of the given function.

Domain of the function

Domain = ( 0,125 ] Domain = ( 0,125 ]

Problem : Find range of the function :

f x = e x − e - | x | e x + e | x | f x = e x − e - | x | e x + e | x |

Solution : We observe that for x≤0,

f x = 0 f x = 0

For x>0

⇒ y = f x = e x − e - | x | e x + e | x | = e 2 x - 1 2 e 2 x ⇒ y = f x = e x − e - | x | e x + e | x | = e 2 x - 1 2 e 2 x

⇒ y X 2 e 2 x = e 2 x - 1 ⇒ 1 - 2 y e 2 x = 1 ⇒ y X 2 e 2 x = e 2 x - 1 ⇒ 1 - 2 y e 2 x = 1 ⇒ e 2 x = 1 1 − 2 y ⇒ e 2 x = 1 1 − 2 y

We can see that e 2 x ≥ 1 e 2 x ≥ 1 for all x. Hence,

⇒ 1 1 − 2 y ≥ 1 ⇒ 1 1 − 2 y − 1 ≥ 0 ⇒ 1 − 1 + 2 y 1 − 2 y ≥ 0 ⇒ 1 1 − 2 y ≥ 1 ⇒ 1 1 − 2 y − 1 ≥ 0 ⇒ 1 − 1 + 2 y 1 − 2 y ≥ 0

⇒ 2 y 1 − 2 y ≥ 0 ⇒ 2 y 1 − 2 y ≥ 0 ⇒ 2 y 2 y − 1 ≤ 0 ⇒ 2 y 2 y − 1 ≤ 0

Here, critical points are 0,1. Thus, range of the given function is :

Range = [ 0, 1 2 ) Range = [ 0, 1 2 )