To solve each part, we will evaluate the functions as required by the operations. Let's go through each part step-by-step.

Given:

• $f(x) = \frac{1}{x + 2}$
• $g(x) = 4x - 2$
• $h(x) = -x^2 + 9$
• $j(x) = |x|$

(a) $(g + h)(3)$

To find $(g + h)(3)$, we calculate $g(3)$ and $h(3)$, then add the results.

1. Calculate $g(3)$: $g(3) = 4 \cdot 3 - 2 = 12 - 2 = 10$

2. Calculate $h(3)$: $h(3) = -(3)^2 + 9 = -9 + 9 = 0$

3. Add $g(3)$ and $h(3)$: $(g + h)(3) = 10 + 0 = 10$

So, $(g + h)(3) = 10$.

(b) $(j - f)(-3)$

To find $(j - f)(-3)$, we calculate $j(-3)$ and $f(-3)$, then subtract $f(-3)$ from $j(-3)$.

1. Calculate $j(-3)$: $j(-3) = |-3| = 3$

2. Calculate $f(-3)$: $f(-3) = \frac{1}{-3 + 2} = \frac{1}{-1} = -1$

3. Subtract $f(-3)$ from $j(-3)$: $(j - f)(-3) = 3 - (-1) = 3 + 1 = 4$

So, $(j - f)(-3) = 4$.

(c) $(hg)(-8)$

To find $(hg)(-8)$, we calculate $h(-8)$ and $g(-8)$, then multiply the results.

1. Calculate $h(-8)$: $h(-8) = -(-8)^2 + 9 = -64 + 9 = -55$

2. Calculate $g(-8)$: $g(-8) = 4 \cdot (-8) - 2 = -32 - 2 = -34$

3. Multiply $h(-8)$ and $g(-8)$: $(hg)(-8) = (-55) \cdot (-34) = 1870$

So, $(hg)(-8) = 1870$.

(d) $\frac{j}{f}(4)$

To find $\frac{j}{f}(4)$, we calculate $j(4)$ and $f(4)$, then divide $j(4)$ by $f(4)$.

1. Calculate $j(4)$: $j(4) = |4| = 4$

2. Calculate $f(4)$: $f(4) = \frac{1}{4 + 2} = \frac{1}{6}$

3. Divide $j(4)$ by $f(4)$: $\frac{j}{f}(4) = \frac{4}{\frac{1}{6}} = 4 \cdot 6 = 24$

So, $\frac{j}{f}(4) = 24$.

Final Answers:

• (a) $(g + h)(3) = 10$
• (b) $(j - f)(-3) = 4$
• (c) $(hg)(-8) = 1870$
• (d) $\frac{j}{f}(4) = 24$

Would you like further details or have any questions?

Related Questions:

1. How would the solution change if $g(x) = 4x + 2$ instead of $4x - 2$?
2. What would $(g \cdot f)(3)$ be given these functions?
3. Can $f(x)$ ever equal zero? If so, for what value of $x$?
4. How does changing the sign in $h(x) = x^2 + 9$ affect the answer for $(hg)(-8)$?
5. What would be the result of $(j + h)(3)$?

Tip:

Always verify domain restrictions, particularly with rational functions like $f(x) = \frac{1}{x + 2}$, to avoid division by zero errors.