Question

for the function $f(x) = \sqrt{x^2 + 100}$, find $f(x)$. then find $f(0)$ and $f(9)$. to find $f(x)$, first find $f(x)$. $f(x) = \square$

Explanation:

Step1: Find first derivative f'(x)

Rewrite \( f(x) = (x^2 + 100)^{1/2} \). Use chain rule: \( f'(x) = \frac{1}{2}(x^2 + 100)^{-1/2} \cdot 2x = \frac{x}{(x^2 + 100)^{1/2}} \)

Step2: Find second derivative f''(x)

Use quotient rule: \( f''(x) = \frac{(1)(x^2 + 100)^{1/2} - x \cdot \frac{1}{2}(x^2 + 100)^{-1/2} \cdot 2x}{x^2 + 100} = \frac{(x^2 + 100) - x^2}{(x^2 + 100)^{3/2}} = \frac{100}{(x^2 + 100)^{3/2}} \)

Step3: Calculate f''(0)

Substitute x=0: \( f''(0) = \frac{100}{(0 + 100)^{3/2}} = \frac{100}{1000} = 0.1 \)

Step4: Calculate f''(9)

Substitute x=9: \( x^2 + 100 = 81 + 100 = 181 \), so \( f''(9) = \frac{100}{181^{3/2}} = \frac{100}{181\sqrt{181}} \approx \frac{100}{181 \cdot 13.4536} \approx 0.040 \)

Answer:

First derivative \( f'(x) = \frac{x}{\sqrt{x^2 + 100}} \); Second derivative \( f''(x) = \frac{100}{(x^2 + 100)^{3/2}} \); \( f''(0) = 0.1 \); \( f''(9) = \frac{100}{181\sqrt{181}} \) (or approximately 0.040)