Question

21. If \(f\left( x \right) = {5^x}\), show that

\(\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = {5^x}\left( {\frac{{{5^h} - 1}}{h}} \right)\)

Short answer

It has been shown that \(\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = {5^x}\left( {\frac{{{5^h} - {5^x}}}{h}} \right)\).

Step-by-step solution

Obtain the function \(f\left( {x + h} \right)\)

Substitute \(x + h\) for \(x\) into \(f\left( x \right)\) and simplify to get the function \(f\left( {x + h} \right)\) as shown below:

\(\begin{aligned}f\left( {x + h} \right) &= {5^{\left( {x + h} \right)}}\\ &= {5^x}{5^h}\end{aligned}\)

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

Plug the expressions for \(f\left( {x + h} \right)\) and \(f\left( x \right)\) into \(\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h}\) and simplify to obtain the required expression as shown below:

\(\begin{aligned}\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} &= \frac{{{5^x}{5^h} - {5^x}}}{h}\\ &= {5^x}\left( {\frac{{{5^h} - {5^x}}}{h}} \right)\end{aligned}\)

So, \(\frac{{f\left( {x + h} \right) - f\left( x \right)}}{h} = {5^x}\left( {\frac{{{5^h} - {5^x}}}{h}} \right)\).