Question

Evaluate the difference quotient for the given function. Simplify your answer.

38. \(f\left( x \right) = \sqrt {x + 2} \), \(\frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}\)

Short answer

The expression is evaluated as \(\frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \frac{1}{{\sqrt {x - 2} + \sqrt 3 }}\).

Step-by-step solution

Obtain the function \(f\left( 1 \right)\)

Substitute 1 for \(x\) in the function \(f\left( x \right)\) to get the function \(f\left( 1 \right)\) as shown below:

\(\begin{aligned}f\left( 1 \right) &= \sqrt {1 + 2} \\ &= \sqrt 3 \end{aligned}\)

Obtain the expression for \(\frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}\)

Plug the expressions of \(f\left( x \right)\) and \(f\left( 1 \right)\) into \(\frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}}\) and simplify as shown below:

\(\frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} = \frac{{\sqrt {x + 2} - \sqrt 3 }}{{x - 1}}\)

Rationalize the above function as shown below:

\(\begin{aligned}\frac{{f\left( x \right) - f\left( 1 \right)}}{{x - 1}} &= \frac{{\left( {\sqrt {x + 2} - \sqrt 3 } \right)}}{{x - 1}} \cdot \frac{{\left( {\sqrt {x + 2} + \sqrt 3 } \right)}}{{\left( {\sqrt {x + 2} + \sqrt 3 } \right)}}\\ &= \frac{{\left( {x + 2} \right) - 3}}{{\left( {x - 1} \right)\left( {\sqrt {x - 2} + \sqrt 3 } \right)}}\\ &= \frac{{x - 1}}{{\left( {x - 1} \right)\left( {\sqrt {x - 2} + \sqrt 3 } \right)}}\\ &= \frac{1}{{\sqrt {x - 2} + \sqrt 3 }}\end{aligned}\)

The expression \(\frac{1}{{\sqrt {x - 2} + \sqrt 3 }}\)cannot be simplified further. So, it is considered as final answer.