1. **Problem 1: Draw the graph of the piecewise function and find its range** Given: $$f(x) = \begin{cases} 1 + 2x, & x < 0 \\ 3 + 5x, & x \geq 0 \end{cases}$$ 2. **Step 1: Understand the function** - For $x < 0$, the function is linear with slope 2 and intercept 1. - For $x \geq 0$, the function is linear with slope 5 and intercept 3. 3. **Step 2: Find the value at the boundary $x=0$** - From left: $f(0^-) = 1 + 2 \times 0 = 1$ - From right: $f(0) = 3 + 5 \times 0 = 3$ The function has a jump discontinuity at $x=0$. 4. **Step 3: Find the range for each piece** - For $x < 0$, as $x \to -\infty$, $1 + 2x \to -\infty$ and as $x \to 0^-$, $f(x) \to 1$ from below. - So range for $x<0$ is $(-\infty, 1)$. - For $x \geq 0$, as $x \to 0$, $f(x) = 3$, and as $x \to +\infty$, $3 + 5x \to +\infty$. - So range for $x \geq 0$ is $[3, +\infty)$. 5. **Step 4: Combine ranges** - The total range is $(-\infty, 1) \cup [3, +\infty)$. --- 6. **Problem 2: Prove that $[f(x)]^2 = f(x^3) + 3f(1/x)$ for $f(x) = x - \frac{1}{x}$** 7. **Step 1: Write down the function and expressions** $$f(x) = x - \frac{1}{x}$$ Calculate $[f(x)]^2$: $$[f(x)]^2 = \left(x - \frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2}$$ 8. **Step 2: Calculate $f(x^3)$** $$f(x^3) = x^3 - \frac{1}{x^3}$$ 9. **Step 3: Calculate $f(1/x)$** $$f\left(\frac{1}{x}\right) = \frac{1}{x} - x = -\left(x - \frac{1}{x}\right) = -f(x)$$ 10. **Step 4: Calculate $f(x^3) + 3f(1/x)$** $$f(x^3) + 3f\left(\frac{1}{x}\right) = x^3 - \frac{1}{x^3} + 3\left(-f(x)\right) = x^3 - \frac{1}{x^3} - 3\left(x - \frac{1}{x}\right)$$ Simplify the right side: $$= x^3 - \frac{1}{x^3} - 3x + \frac{3}{x}$$ 11. **Step 5: Express $x^3 - \frac{1}{x^3}$ in terms of $x - \frac{1}{x}$** Recall the identity: $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ Set $a = x$, $b = \frac{1}{x}$: $$x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right)\left(x^2 + x \cdot \frac{1}{x} + \frac{1}{x^2}\right) = \left(x - \frac{1}{x}\right)(x^2 + 1 + \frac{1}{x^2})$$ 12. **Step 6: Substitute back** $$f(x^3) + 3f\left(\frac{1}{x}\right) = \left(x - \frac{1}{x}\right)(x^2 + 1 + \frac{1}{x^2}) - 3\left(x - \frac{1}{x}\right) = \left(x - \frac{1}{x}\right)(x^2 + 1 + \frac{1}{x^2} - 3)$$ Simplify inside the parenthesis: $$x^2 + 1 + \frac{1}{x^2} - 3 = x^2 - 2 + \frac{1}{x^2}$$ 13. **Step 7: Final expression** $$f(x^3) + 3f\left(\frac{1}{x}\right) = \left(x - \frac{1}{x}\right)(x^2 - 2 + \frac{1}{x^2})$$ Recall from Step 7: $$[f(x)]^2 = x^2 - 2 + \frac{1}{x^2}$$ Therefore: $$[f(x)]^2 = \left(x - \frac{1}{x}\right)^2 = \left(x - \frac{1}{x}\right)(x^2 - 2 + \frac{1}{x^2}) = f(x^3) + 3f\left(\frac{1}{x}\right)$$ **Hence proved.**