Question

the limit represents the derivative of some function f at some number a. state such an f and a.

$$\lim_{x \to 5} \frac{2^x - 32}{x - 5}$$

$$\bigcirc f(x) = 5^x, a = 2$$

$$\bigcirc f(x) = 2^x, a = 32$$

$$\bigcirc f(x) = 2^x, a = 5$$

$$\bigcirc f(x) = x^2, a = 5$$

$$\bigcirc f(x) = 5^x, a = 32$$

Explanation:

Step1: Recall the definition of the derivative

The derivative of a function \(f(x)\) at \(x = a\) is given by \(f^{\prime}(a)=\lim_{x
ightarrow a}\frac{f(x)-f(a)}{x - a}\).

Step2: Compare with the given limit

We are given \(\lim_{x
ightarrow5}\frac{2^{x}-32}{x - 5}\). Since \(32=2^{5}\), we can rewrite the limit as \(\lim_{x
ightarrow5}\frac{2^{x}-2^{5}}{x - 5}\).
Comparing with \(f^{\prime}(a)=\lim_{x
ightarrow a}\frac{f(x)-f(a)}{x - a}\), we have \(f(x)=2^{x}\) and \(a = 5\).

Answer:

\(f(x)=2^{x}\), \(a = 5\) (the third option)