Question

let \\(f(x) = \frac{9}{\sqrt{x}}\\).

step 1 of 2: use the definition of the derivative at a point to find \\(f(16)\\).

answer

\\(f(16) =\\)

Explanation:

State the definition of the derivative at a point

The derivative of a function \(f(x)\) at a point \(x = a\) is defined by the limit:

$$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a}$$

For this problem, we have:

$$f(x) = \frac{9}{\sqrt{x}}, \quad a = 16$$

Evaluate the function at the given point

Calculate the value of \(f(16)\):

$$f(16) = \frac{9}{\sqrt{16}} = \frac{9}{4}$$

Set up the limit expression

Substitute \(f(x)\), \(f(16)\), and \(a = 16\) into the limit definition:

$$f'(16) = \lim_{x \to 16} \frac{\frac{9}{\sqrt{x}} - \frac{9}{4}}{x - 16}$$

Simplify the numerator of the limit

Find a common denominator for the terms in the numerator:

$$\frac{9}{\sqrt{x}} - \frac{9}{4} = \frac{36 - 9\sqrt{x}}{4\sqrt{x}} = \frac{9(4 - \sqrt{x})}{4\sqrt{x}}$$

Substitute this back into the limit expression:

$$f'(16) = \lim_{x \to 16} \frac{9(4 - \sqrt{x})}{4\sqrt{x}(x - 16)}$$

Rationalize and evaluate the limit

Factor the denominator using the difference of squares, \(x - 16 = (\sqrt{x} - 4)(\sqrt{x} + 4)\):

$$f'(16) = \lim_{x \to 16} \frac{-9(\sqrt{x} - 4)}{4\sqrt{x}(\sqrt{x} - 4)(\sqrt{x} + 4)}$$

Cancel the common factor \((\sqrt{x} - 4)\) for \(x
eq 16\):

$$f'(16) = \lim_{x \to 16} \frac{-9}{4\sqrt{x}(\sqrt{x} + 4)}$$

Evaluate the limit by direct substitution:

$$f'(16) = \frac{-9}{4\sqrt{16}(\sqrt{16} + 4)} = \frac{-9}{4(4)(4 + 4)} = \frac{-9}{16(8)} = -\frac{9}{128}$$

Answer:

Let \(f(x) = \frac{9}{\sqrt{x}}\). Use the definition of the derivative at a point to find \(f'(16)\).

\(f'(16) =\) <blank>\(-\frac{9}{128}\)</blank>