EVALUATING EXPONENTIAL FUNCTIONS WORKSHEET

Problem 1 : 

Evaluate :

f(x)  =  (1/12) ⋅ 6^x at x = 3

Problem 2 : 

Evaluate :

f(x)  =  5 ⋅ 3ⁿ at n = 4

Problem 3 : 

Evaluate :

f(n)  =  8 ⋅ 2ⁿ at n = -3

Problem 4 : 

Evaluate :

g(x)  =  (1/2) ⋅ (1/3)^x at x = 3

Problem 5 : 

Evaluate :

g(x)  =  (1/243) ⋅ 3^x at x = 4

Problem 6 : 

The approximate number of Calories C that an animal needs each day is given by C = 72m^3/5, where m is the mass of animal in kilograms. Find the number of Calories that a 243 kg animal needs each day.   

Detailed Answer Key

Problem 1 : 

Evaluate :

f(x)  =  (1/12) ⋅ 6^x at x = 3

Solution : 

f(x)  =  (1/12) ⋅ 6^x

Substitute 3 for x. 

f(3)  =  (1/12) ⋅ 6³

=  (1/12) ⋅ 216

=  18

Problem 2 : 

Evaluate :

f(x)  =  5 ⋅ 3ⁿ at n = 4

Solution : 

f(x)  =  5 ⋅ 3ⁿ

Substitute 4 for n. 

f(4)  =  5 ⋅ 3⁴

=  5 ⋅ 81

=  405

Problem 3 : 

Evaluate :

f(n)  =  8 ⋅ 2ⁿ at n = -3

Solution : 

f(n)  =  8 ⋅ 2ⁿ

Substitute -3 for n. 

f(-3)  =  8 ⋅ 2^-3

=  8 ⋅ 1/2³

=  8 ⋅ 1/8

=  1

Problem 4 : 

Evaluate :

g(x)  =  (1/2) ⋅ (1/3)^x at x = 3

Solution : 

g(x)  =  (1/2) ⋅ (1/3)^x

Substitute 3 for x. 

g(3)  =  (1/2) ⋅ (1/3)³

Power of a Quotient Property. 

=  (1/2) ⋅ (1³/3³)

=  (1/2) ⋅ (1/27)

=  1/54

Problem 5 : 

Evaluate :

g(x)  =  (1/243) ⋅ 3^x at x = 4

Solution : 

g(x)  =  (1/243) ⋅ 3^x

Substitute 4 for x. 

g(4)  =  (1/243) ⋅ 3⁴

Power of a Quotient Property. 

=  (1/243) ⋅ 81

=  81/243

=  1/3

Problem 6 : 

The approximate number of Calories C that an animal needs each day is given by C = 72m^3/5, where m is the mass of animal in kilograms. Find the number of Calories that a 243 kg animal needs each day.  

Solution : 

C  =  72m^3/5

Substitute 243 for m. 

=  72(243)^3/5

=  72(3⁵)^3/5

Power of a Power Property.

=  72(3^{5 ⋅ 3/5})

=  72(3³)

=  72(27)

=  1944

The animal needs 1944 Calories per day to maintain health. 

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