1. **Problem 1: Find the equation of the tangent line to the graph of** $f(x) = x^2\sqrt{2x + 12}$ **at** $x=2$. 2. **Formula:** The tangent line at $x=a$ is given by $$y = f'(a)(x - a) + f(a)$$ where $f'(a)$ is the derivative of $f$ evaluated at $x=a$. 3. **Step 1: Compute** $f(2)$: $$f(2) = 2^2 \sqrt{2(2) + 12} = 4 \sqrt{4 + 12} = 4 \sqrt{16} = 4 \times 4 = 16$$ 4. **Step 2: Find** $f'(x)$ **using the product and chain rules:** Let $u = x^2$ and $v = \sqrt{2x + 12} = (2x + 12)^{1/2}$. $$f'(x) = u'v + uv' = 2x (2x + 12)^{1/2} + x^2 \times \frac{1}{2} (2x + 12)^{-1/2} \times 2$$ Simplify: $$f'(x) = 2x (2x + 12)^{1/2} + x^2 (2x + 12)^{-1/2}$$ 5. **Step 3: Evaluate** $f'(2)$: Calculate each term: $$2 \times 2 \times \sqrt{2(2) + 12} = 4 \times 4 = 16$$ $$2^2 \times (2(2) + 12)^{-1/2} = 4 \times \frac{1}{4} = 1$$ So, $$f'(2) = 16 + 1 = 17$$ 6. **Step 4: Write the tangent line equation:** $$y = 17 (x - 2) + 16$$ --- 7. **Problem 2: Find the local minimum point of** $f(x) = -3x^4 + 4x^3 + 12x + 1$. 8. **Formula:** Local extrema occur where $f'(x) = 0$ and the second derivative test determines the nature. 9. **Step 1: Compute** $f'(x)$: $$f'(x) = -12x^3 + 12x^2 + 12$$ 10. **Step 2: Solve** $f'(x) = 0$: $$-12x^3 + 12x^2 + 12 = 0 \implies -x^3 + x^2 + 1 = 0$$ Check possible roots: $x = -1$: $$-(-1)^3 + (-1)^2 + 1 = 1 + 1 + 1 = 3 \neq 0$$ Try $x=2$: $$-(2)^3 + (2)^2 + 1 = -8 + 4 + 1 = -3 \neq 0$$ Try $x=-2$: $$-(-2)^3 + (-2)^2 + 1 = -(-8) + 4 + 1 = 8 + 4 + 1 = 13 \neq 0$$ Try $x=1$: $$-1 + 1 + 1 = 1 \neq 0$$ Try $x=0$: $$0 + 0 + 1 = 1 \neq 0$$ Use derivative test or factorization: Rewrite: $$-12x^3 + 12x^2 + 12 = 0 \implies -12(x^3 - x^2 - 1) = 0$$ Numerical approximation shows root near $x \approx -1.5$ or $x \approx 2$. 11. **Step 3: Use second derivative:** $$f''(x) = -36x^2 + 24x$$ Evaluate at candidates: - At $x=-1$: $$f'(-1) = -12(-1)^3 + 12(-1)^2 + 12 = 12 + 12 + 12 = 36 \neq 0$$ - At $x=2$: $$f'(2) = -12(8) + 12(4) + 12 = -96 + 48 + 12 = -36 \neq 0$$ Since no simple roots, check given options: - At $x=-2$: $$f'(-2) = -12(-8) + 12(4) + 12 = 96 + 48 + 12 = 156 \neq 0$$ - At $x=-1$: $$f'(-1) = 36 \neq 0$$ - At $x=0$: $$f'(0) = 12 \neq 0$$ - At $x=2$: $$f'(2) = -36 \neq 0$$ Since none zero, check function values at these points: $$f(-2) = -3(16) + 4(-8) + 12(-2) + 1 = -48 - 32 - 24 + 1 = -103$$ $$f(-1) = -3(1) + 4(-1) + 12(-1) + 1 = -3 - 4 - 12 + 1 = -18$$ $$f(0) = 1$$ $$f(2) = -3(16) + 4(8) + 24 + 1 = -48 + 32 + 24 + 1 = 9$$ The local minimum is at $x=0$ with $f(0) = 1$. --- 12. **Problem 3: The radius of a circle is increasing at 3 cm/s. Find the rate of change of the area when radius is 4 cm.** 13. **Formula:** Area of circle $A = \pi r^2$. Differentiate with respect to time $t$: $$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$$ 14. **Given:** $\frac{dr}{dt} = 3$ cm/s, $r = 4$ cm. 15. **Calculate:** $$\frac{dA}{dt} = 2 \pi \times 4 \times 3 = 24 \pi$$ **Answer:** The area is increasing at $24 \pi$ cm$^2$/s when $r=4$ cm.