1. **Problem statement:** We have the function $f(x) = x^3 - 2x^2 - 4x + 8$. We need to calculate $f(x)$ for $x = -3, -1, 1, 4$, show that $f(x)$ can be factored as $f(x) = (x^2 - 4)(x - 2)$, find the zeros of $f$, and determine where $f$ is positive or negative. 2. **Calculate function values:** - $f(-3) = (-3)^3 - 2(-3)^2 - 4(-3) + 8 = -27 - 18 + 12 + 8 = -25$ - $f(-1) = (-1)^3 - 2(-1)^2 - 4(-1) + 8 = -1 - 2 + 4 + 8 = 9$ - $f(1) = 1^3 - 2(1)^2 - 4(1) + 8 = 1 - 2 - 4 + 8 = 3$ - $f(4) = 4^3 - 2(4)^2 - 4(4) + 8 = 64 - 32 - 16 + 8 = 24$ 3. **Factorization:** Start with $f(x) = x^3 - 2x^2 - 4x + 8$. Group terms: $f(x) = (x^3 - 2x^2) + (-4x + 8)$. Factor each group: $x^2(x - 2) - 4(x - 2)$. Factor out $(x - 2)$: $f(x) = (x - 2)(x^2 - 4)$. Note that $x^2 - 4$ is a difference of squares: $x^2 - 4 = (x - 2)(x + 2)$. So $f(x) = (x - 2)(x - 2)(x + 2) = (x - 2)^2 (x + 2)$. 4. **Find zeros:** Set $f(x) = 0$: $$(x - 2)^2 (x + 2) = 0$$ Zeros are $x = 2$ (with multiplicity 2) and $x = -2$. 5. **Sign of $f(x)$:** - For $x < -2$, choose $x = -3$: $f(-3) = -25 < 0$ so $f$ is negative. - For $-2 < x < 2$, choose $x = 0$: $f(0) = 8 > 0$ so $f$ is positive. - For $x > 2$, choose $x = 3$: $f(3) = (3-2)^2(3+2) = 1^2 imes 5 = 5 > 0$ so $f$ is positive. 6. **Find derivative $f'(x)$:** $$f'(x) = 3x^2 - 4x - 4$$ 7. **Find critical points:** Set $f'(x) = 0$: $$3x^2 - 4x - 4 = 0$$ Use quadratic formula: $$x = \frac{4 \pm \sqrt{(-4)^2 - 4 \times 3 \times (-4)}}{2 \times 3} = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm \sqrt{64}}{6} = \frac{4 \pm 8}{6}$$ So, $$x_1 = \frac{4 - 8}{6} = \frac{-4}{6} = -\frac{2}{3}$$ $$x_2 = \frac{4 + 8}{6} = \frac{12}{6} = 2$$ 8. **Determine increasing/decreasing intervals:** - For $x < -\frac{2}{3}$, test $f'(-1) = 3(1) + 4 - 4 = 3 - 4 - 4 = -5 < 0$, so $f$ is decreasing. - For $-\frac{2}{3} < x < 2$, test $f'(0) = -4 < 0$, so $f$ is decreasing. - For $x > 2$, test $f'(3) = 3(9) - 12 - 4 = 27 - 16 = 11 > 0$, so $f$ is increasing. 9. **Local extrema:** - At $x = -\frac{2}{3}$, $f'(x)$ changes from negative to negative (no sign change), so no local extremum. - At $x = 2$, $f'(x)$ changes from negative to positive, so local minimum at $x=2$. Calculate $f(2)$: $$f(2) = (2 - 2)^2 (2 + 2) = 0$$ 10. **Global extrema:** Since $f(x) \to \infty$ as $x \to \infty$ and $f(x) \to -\infty$ as $x \to -\infty$, the local minimum at $x=2$ with $f(2)=0$ is a global minimum. Final answers: - $f(-3) = -25$, $f(-1) = 9$, $f(1) = 3$, $f(4) = 24$ - $f(x) = (x - 2)^2 (x + 2)$ - Zeros at $x = -2$ and $x = 2$ - $f$ negative for $x < -2$, positive for $x > -2$ - $f'(x) = 3x^2 - 4x - 4$ - $f$ decreasing on $(-\infty, 2)$, increasing on $(2, \infty)$ - Local and global minimum at $x=2$, $f(2)=0$