1. **Problem 1:** Find the derivative $f'(x)$ if $f(x) = (x^3 - x)(x^2 - 2)$. - Use the product rule: $\frac{d}{dx}[u \cdot v] = u'v + uv'$. - Let $u = x^3 - x$ and $v = x^2 - 2$. - Compute derivatives: $u' = 3x^2 - 1$, $v' = 2x$. - Apply product rule: $$f'(x) = (3x^2 - 1)(x^2 - 2) + (x^3 - x)(2x)$$ - Expand terms: $= (3x^2 - 1)(x^2 - 2) + 2x(x^3 - x)$ $= (3x^2)(x^2) - (3x^2)(2) - 1(x^2) + 1(2) + 2x^4 - 2x^2$ $= 3x^4 - 6x^2 - x^2 + 2 + 2x^4 - 2x^2$ - Combine like terms: $= (3x^4 + 2x^4) + (-6x^2 - x^2 - 2x^2) + 2$ $= 5x^4 - 9x^2 + 2$ - **Answer:** $f'(x) = 5x^4 - 9x^2 + 2$. 2. **Problem 2:** Find the equation of the tangent line to $y = 3x^2 + 2x + 1$ with slope $-4$. - The derivative gives slope: $y' = 6x + 2$. - Set slope equal to $-4$: $$6x + 2 = -4$$ $$6x = -6$$ $$x = -1$$ - Find $y$ at $x = -1$: $$y = 3(-1)^2 + 2(-1) + 1 = 3 - 2 + 1 = 2$$ - Equation of tangent line: $$y - y_1 = m(x - x_1)$$ $$y - 2 = -4(x + 1)$$ $$y = -4x - 4 + 2 = -4x - 2$$ - None of the options exactly match $y = -4x - 2$, but closest is $y = -4x + 6$ (incorrect). - Recheck: Options given are $y=4x$, $y=4x+1$, $y=-4x-4$, $y=-4x+6$. - Our derived line is $y = -4x - 2$, which is not listed. Possibly a typo in options. The closest is $y = -4x - 4$. - **Answer:** $y = -4x - 4$ (closest match). 3. **Problem 3:** Find the instantaneous rate of change of $f(x) = 3x^2 - 6x$ at $x=1$. - Derivative: $$f'(x) = 6x - 6$$ - Evaluate at $x=1$: $$f'(1) = 6(1) - 6 = 0$$ - None of the options match 0. Recheck options: 15, 12, -12, -15. - Possibly a mistake in problem or options. Confirm derivative: $6x - 6$. At $x=1$, $6 - 6 = 0$. - If problem intended $f(x) = 3x^2 + 6x$, then derivative $6x + 6$, at $x=1$ is 12. - Assuming typo, answer is 0, but closest option is 12. - **Answer:** 0 (not listed), closest is 12. **Summary:** - $f'(x) = 5x^4 - 9x^2 + 2$ - Tangent line: $y = -4x - 4$ (closest option) - Instantaneous rate of change at $x=1$ is 0 (not listed)