Q. 118 - Precalculus Enhanced with Graphing Utilities

Question

If $f\left(x\right)=a^x$, show that  $\frac{f\left(x+h\right)-f\left(x\right)}{h}=a^x.\frac{a^h-1}{h} h\ne 0$.

Step-by-Step Solution

Answer

To prove $\frac{f\left(x+h\right)-f\left(x\right)}{h}=a^x.\frac{a^h-1}{h}$, first substitute the values of $f\left(x\right),f\left(x+h\right)$ and solve to get the expression at the RHS.

Step 1. Given information.

Consider the given question,

$$\begin{gathered} f\left(x\right)=a^x, \\ \frac{f\left(x+h\right)-f\left(x\right)}{h}=a^x.\frac{a^h-1}{h} h\ne 0 \end{gathered}$$

Substitute  $$\begin{gathered} f\left(x\right)=a^x, \\ f\left(x+h\right)=a^{x+h} \end{gathered}$$ in the given expression,

Take the LHS,

$\frac{f\left(x+h\right)-f\left(x\right)}{h}=\frac{a^{x+h}-a^x}{h}$

Step 2. Use laws of exponents.

Using laws of exponents, $x^{a.b}=x^a+x^b$,

$$\begin{gathered} =\frac{a^x.a^h-a^{x^{}}}{h} \\ =a^x.\frac{a^h-1}{h} \end{gathered}$$

Therefore, $LHS=RHS$.

Hence, proved.