Question

question 5
given $f(x)=3\sqrt{x - 8}$, find $f(x)$ using the limit definition of the derivative.
$f(x)=$
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Explanation:

Step1: Recall limit definition of derivative

The limit definition of the derivative is:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$

Step2: Substitute $f(x+h)$ and $f(x)$

Substitute $f(x+h)=3\sqrt{x+h - 8}$ and $f(x)=3\sqrt{x - 8}$:
$$f'(x) = \lim_{h \to 0} \frac{3\sqrt{x+h - 8} - 3\sqrt{x - 8}}{h}$$
Factor out the 3:
$$f'(x) = 3\lim_{h \to 0} \frac{\sqrt{x+h - 8} - \sqrt{x - 8}}{h}$$

Step3: Rationalize the numerator

Multiply numerator and denominator by the conjugate $\sqrt{x+h - 8} + \sqrt{x - 8}$:
$$f'(x) = 3\lim_{h \to 0} \frac{(\sqrt{x+h - 8} - \sqrt{x - 8})(\sqrt{x+h - 8} + \sqrt{x - 8})}{h(\sqrt{x+h - 8} + \sqrt{x - 8})}$$
Use the difference of squares $(a-b)(a+b)=a^2-b^2$:
$$f'(x) = 3\lim_{h \to 0} \frac{(x+h - 8) - (x - 8)}{h(\sqrt{x+h - 8} + \sqrt{x - 8})}$$

Step4: Simplify the numerator

Simplify the terms in the numerator:
$$f'(x) = 3\lim_{h \to 0} \frac{x+h - 8 - x + 8}{h(\sqrt{x+h - 8} + \sqrt{x - 8})} = 3\lim_{h \to 0} \frac{h}{h(\sqrt{x+h - 8} + \sqrt{x - 8})}$$
Cancel $h$ in numerator and denominator:
$$f'(x) = 3\lim_{h \to 0} \frac{1}{\sqrt{x+h - 8} + \sqrt{x - 8}}$$

Step5: Evaluate the limit

Substitute $h=0$ into the expression:
$$f'(x) = 3 \cdot \frac{1}{\sqrt{x - 8} + \sqrt{x - 8}} = 3 \cdot \frac{1}{2\sqrt{x - 8}}$$

Answer:

$\frac{3}{2\sqrt{x-8}}$