If $a$ is a positive real number, find the following values for the given function.
$g(x)=2 x-5$
(a) $g(\frac{1}{a})$
(b) $\frac{1}{g(a)}$
(c) $g(\sqrt{a})$
(d) $\sqrt{g(a)}$
Answer (1)
Substitute a 1 into g ( x ) to find g ( a 1 ) = a 2 − 5 .
Substitute a into g ( x ) and find the reciprocal to get g ( a ) 1 = 2 a − 5 1 .
Substitute a into g ( x ) to find g ( a ) = 2 a − 5 .
Substitute a into g ( x ) and take the square root to get g ( a ) = 2 a − 5 .
Explanation
Understanding the Problem We are given the function g ( x ) = 2 x − 5 , and we need to find the expressions for (a) g ( a 1 ) , (b) g ( a ) 1 , (c) g ( a ) , and (d) g ( a ) .
Finding g(1/a) (a) To find g ( a 1 ) , we substitute a 1 for x in the expression for g ( x ) : g ( a 1 ) = 2 ( a 1 ) − 5 = a 2 − 5.
Finding 1/g(a) (b) To find g ( a ) 1 , we first find g ( a ) by substituting a for x in the expression for g ( x ) : g ( a ) = 2 a − 5.
Then, we take the reciprocal: g ( a ) 1 = 2 a − 5 1 .
Finding g(√a) (c) To find g ( a ) , we substitute a for x in the expression for g ( x ) : g ( a ) = 2 a − 5.
Finding √g(a) (d) To find g ( a ) , we first find g ( a ) by substituting a for x in the expression for g ( x ) : g ( a ) = 2 a − 5.
Then, we take the square root: g ( a ) = 2 a − 5 .
Final Answer Therefore, the expressions are: (a) g ( a 1 ) = a 2 − 5 (b) g ( a ) 1 = 2 a − 5 1 (c) g ( a ) = 2 a − 5 (d) g ( a ) = 2 a − 5
Examples
Understanding function composition and variable substitution is crucial in many real-world applications. For example, in physics, if g ( x ) represents the velocity of an object at time x , then g ( a 1 ) could represent the velocity at a specific fraction of time. Similarly, in economics, if g ( a ) represents the profit from selling a units of a product, then g ( a ) 1 might represent the cost per unit of profit. These substitutions help analyze and optimize various processes by relating different variables through functions.
