1. **Problem 1:** For what number $k$ is $x+1$ a factor of $f(x) = 2x^6 - kx^3 + 5x - 1$? By the Factor Theorem, if $x+1$ is a factor, then $f(-1) = 0$. Calculate $f(-1)$: $$f(-1) = 2(-1)^6 - k(-1)^3 + 5(-1) - 1 = 2(1) + k - 5 - 1 = 2 + k - 6 = k - 4$$ Set equal to zero: $$k - 4 = 0 \implies k = 4$$ --- 2. **Problem 2:** Given $f(x) = 2x^3 - \frac{1}{x}$, which of the following is NOT true? First, find the derivative $f'(x)$: $$f'(x) = \frac{d}{dx}\left(2x^3 - x^{-1}\right) = 6x^2 + x^{-2} = 6x^2 + \frac{1}{x^2}$$ Evaluate at the given points: - $f'(-1) = 6(-1)^2 + \frac{1}{(-1)^2} = 6 + 1 = 7$ (True) - $f'(0)$ is undefined because $\frac{1}{x^2}$ is undefined at $x=0$ (so $f'(0) = 0$ is NOT true) - $f'(1) = 6(1)^2 + \frac{1}{1^2} = 6 + 1 = 7$ (True) - $f'(2) = 6(2)^2 + \frac{1}{2^2} = 6 \times 4 + \frac{1}{4} = 24 + 0.25 = \frac{97}{4}$ (True) So the statement $f'(0) = 0$ is NOT true. --- 3. **Problem 3:** What are the domain and range of $f(x) = 8x^{-3/4}$? Rewrite: $$f(x) = 8 \cdot \frac{1}{x^{3/4}}$$ - The domain requires $x^{3/4}$ to be defined and real. Since the exponent is a fraction with an even denominator (4), $x$ must be positive to keep the function real. - So domain: $(0, \infty)$ - For range, as $x \to 0^+$, $x^{-3/4} \to \infty$, so $f(x) \to \infty$. - As $x \to \infty$, $x^{-3/4} \to 0^+$, so $f(x) \to 0^+$. Thus, range is also $(0, \infty)$. --- **Final answers:** 1. $k = 4$ 2. $f'(0) = 0$ is NOT true 3. Domain and range are $(0, \infty)$ and $(0, \infty)$ respectively.