1. **Problem statement:** (a) Find the domain where the function $f(x) = |x^2 - 9|$ is differentiable and find the formula for $f'(x)$. (b) Plot the graphs of $f$ and $f'$. (d) For $h(x) = |x - 1| + |x + 2|$, find the domain where $h$ is differentiable and find the formula for $h'(x)$. Also plot the graphs of $h$ and $h'$. 18. Given $f(4) = 2$, $g(4) = 5$, $f'(4) = 6$, $g'(4) = -3$, find $h'(4)$ where $h(x) = f(x)g(x)$. --- 2. **Important rules and formulas:** - The absolute value function $|u|$ is differentiable everywhere except where $u=0$. - For $f(x) = |u(x)|$, the derivative is $f'(x) = \frac{u(x)}{|u(x)|} u'(x)$ for $u(x) \neq 0$. - The product rule: if $h(x) = f(x)g(x)$, then $h'(x) = f'(x)g(x) + f(x)g'(x)$. --- 3. **Solution for (a):** - Here, $u(x) = x^2 - 9$. - $u(x) = 0$ when $x^2 - 9 = 0 \Rightarrow x = \pm 3$. - So $f(x)$ is differentiable for all $x \neq \pm 3$. - Compute $u'(x) = 2x$. - For $x \neq \pm 3$, $$ f'(x) = \frac{u(x)}{|u(x)|} u'(x) = \frac{x^2 - 9}{|x^2 - 9|} \cdot 2x. $$ - This means: - For $|x| > 3$, $x^2 - 9 > 0$, so $f'(x) = 2x$. - For $|x| < 3$, $x^2 - 9 < 0$, so $f'(x) = -2x$. --- 4. **Solution for (b):** - The graph of $f(x) = |x^2 - 9|$ looks like a "W" shape with zeros at $x=\pm 3$. - The derivative $f'(x)$ is piecewise: - $f'(x) = 2x$ for $|x| > 3$. - $f'(x) = -2x$ for $|x| < 3$. - At $x=\pm 3$, $f'(x)$ is not defined. --- 5. **Solution for (d):** - $h(x) = |x - 1| + |x + 2|$. - Points where $h$ is not differentiable are where the inside of absolute values are zero: $x=1$ and $x=-2$. - For $x < -2$: - $h(x) = -(x - 1) - (x + 2) = -x + 1 - x - 2 = -2x - 1$. - So $h'(x) = -2$. - For $-2 < x < 1$: - $h(x) = -(x - 1) + (x + 2) = -x + 1 + x + 2 = 3$. - So $h'(x) = 0$. - For $x > 1$: - $h(x) = (x - 1) + (x + 2) = 2x + 1$. - So $h'(x) = 2$. - $h$ is differentiable for all $x$ except $x = -2$ and $x = 1$. --- 6. **Solution for 18:** - Given $h(x) = f(x)g(x)$. - Using product rule: $$ h'(4) = f'(4)g(4) + f(4)g'(4) = 6 \times 5 + 2 \times (-3) = 30 - 6 = 24. $$ --- **Final answers:** - (a) $f$ is differentiable for $x \neq \pm 3$ and $$ f'(x) = \begin{cases} 2x, & |x| > 3 \\ -2x, & |x| < 3 \end{cases} $$ - (d) $h$ is differentiable for $x \neq -2, 1$ and $$ h'(x) = \begin{cases} -2, & x < -2 \\ 0, & -2 < x < 1 \\ 2, & x > 1 \end{cases} $$ - (18) $h'(4) = 24$.