For the following exercises, consider the function \( f(x)=\sqrt{x-4} \) a) Evaluate the expression \( \frac{f(x+h)-f(x)}{h} \) sqrt \( (x+h-4) / \) sart \( (x-4) \) b) Find \( f^{\prime}(x) \). Question 22 of 30 \( \frac{2(\operatorname{sqrt}(x-4)) /(x-4) /}{\pi} a^{\circ} \infty a_{b}^{a}, a^{b} \leq \sqrt{a}, \sin _{4} \)
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To tackle this problem with the function \( f(x)=\sqrt{x-4} \), let's jump right into evaluating the limit expression! For the expression \( \frac{f(x+h)-f(x)}{h} \), we substitute \( f(x+h) = \sqrt{x+h-4} \) and \( f(x) = \sqrt{x-4} \) to get: \[ \frac{\sqrt{x+h-4} - \sqrt{x-4}}{h}. \] To simplify this, multiply the numerator and the denominator by the conjugate \( \sqrt{x+h-4} + \sqrt{x-4} \): \[ \frac{(\sqrt{x+h-4} - \sqrt{x-4})(\sqrt{x+h-4} + \sqrt{x-4})}{h(\sqrt{x+h-4} + \sqrt{x-4})} = \frac{(x+h-4) - (x-4)}{h(\sqrt{x+h-4} + \sqrt{x-4})} = \frac{h}{h(\sqrt{x+h-4} + \sqrt{x-4})} = \frac{1}{\sqrt{x+h-4} + \sqrt{x-4}}. \] Now, taking the limit as \( h \to 0 \), we find: \[ \lim_{h \to 0} \frac{1}{\sqrt{x+h-4} + \sqrt{x-4}} = \frac{1}{2\sqrt{x-4}}. \] Thus, we can proceed with finding \( f'(x) \). For part b, using the limit derived, we can conclude: \[ f'(x) = \frac{1}{2\sqrt{x-4}}. \] Now you have a solid basis for understanding the derivative of the square root function! Imagine ancient mathematicians scratching their heads over the concept of derivatives – they likely would have marveled at how we can quickly find slopes and rates of change with just a few algebraic tricks today! To put it into practice, think of how knowing the slope of a curve can inform real-world scenarios. For example, in physics, this concept helps us understand motion; the velocity is simply the derivative of the position function with respect to time, allowing us to predict how far a car has traveled at any given instance!
