1. Problem 8: Evaluate $(f \circ g)'(1)$ given $f(1)=2$, $g(1)=2$, $f'(1)=3$, $g'(1)=2$, $f'(2)=1$, and $g'(2)=3$. The chain rule states: $$ (f \circ g)'(x) = f'(g(x)) \cdot g'(x) $$ Calculate: $$ (f \circ g)'(1) = f'(g(1)) \cdot g'(1) = f'(2) \cdot 2 = 1 \cdot 2 = 2 $$ Answer: (A) 2 2. Problem 9: An object is dropped and hits the ground after 5 seconds. Find the height from which it was dropped. Given acceleration due to gravity $a(t) = -9.8$ m/s$^2$ (taking upward positive). Use the kinematic equation for displacement: $$ s = ut + \frac{1}{2}at^2 $$ Initial velocity $u=0$ (dropped), time $t=5$ s, acceleration $a = -9.8$ m/s$^2$. Calculate height (displacement downward): $$ s = 0 + \frac{1}{2}(-9.8)(5)^2 = -4.9 \times 25 = -122.5 \text{ m} $$ Height is positive distance, so height = $4.9 \times 25$ m. Answer: (E) 4.9 × (25)m 3. Problem 10: Population grows proportionally to its size. Initial population $P_0=10000$, doubles in 15 years. Find time to triple. Model: $P(t) = P_0 e^{kt}$. Given doubling: $$ 2P_0 = P_0 e^{15k} \Rightarrow 2 = e^{15k} \Rightarrow \ln 2 = 15k \Rightarrow k = \frac{\ln 2}{15} $$ Find $t$ for tripling: $$ 3P_0 = P_0 e^{kt} \Rightarrow 3 = e^{kt} \Rightarrow \ln 3 = kt = \frac{\ln 2}{15} t \Rightarrow t = 15 \frac{\ln 3}{\ln 2} $$ Answer: (D) 15 ln(3)/ln(2) 4. Problem 11: Side of cube increases at 3 cm/s. Find rate of volume increase when side length $s=2$ cm. Volume of cube: $$ V = s^3 $$ Differentiate w.r.t. time $t$: $$ \frac{dV}{dt} = 3s^2 \frac{ds}{dt} $$ Given $\frac{ds}{dt} = 3$ cm/s, $s=2$ cm: $$ \frac{dV}{dt} = 3 \times (2)^2 \times 3 = 3 \times 4 \times 3 = 36 \text{ cm}^3/\text{s} $$ Answer: (E) 36 cm^3/s 5. Problem 12: Find minimum slope of tangent to curve $f(x) = x^5 + x^3 - 2x$. Slope of tangent is derivative: $$ f'(x) = 5x^4 + 3x^2 - 2 $$ Find critical points of $f'(x)$ by differentiating again: $$ f''(x) = 20x^3 + 6x $$ Set $f''(x) = 0$: $$ 20x^3 + 6x = 0 \Rightarrow x(20x^2 + 6) = 0 $$ Solutions: $$ x=0 \quad \text{or} \quad 20x^2 + 6=0 \Rightarrow x^2 = -\frac{6}{20} \text{ (no real roots)} $$ So only critical point at $x=0$. Evaluate $f'(x)$ at $x=0$: $$ f'(0) = -2 $$ Since $f'(x)$ is a quartic with positive leading coefficient, minimum slope is $-2$. Answer: (C) -2