Question: Let \( f(t) = \frac{g(t)}{t^2} \), where \( g(t) = t^3 - 2t^2 + 2e^t \sqrt{t} \). Find \(…

Let \( f(t) = \frac{g(t)}{t^2} \), where \( g(t) = t^3 - 2t^2 + 2e^t \sqrt{t} \). Find

\( g’(1) = \boxed{\phantom{0}} \)

\( f’(1) = \boxed{\phantom{0}} \)

Give your reasons

Solution

To solve the problem, we need to find \( g’(1) \) and \( f’(1) \). First, let’s find \( g’(t) \). Given \( g(t) = t^3 - 2t^2 + 2e^t \sqrt{t} \). Differentiate each term: 1. \( \frac{d}{dt}(t^3) = 3t^2 \) 2. \( \frac{d}{dt}(-2t^2) = -4t \) 3. For \( 2e^t \sqrt{t} \), use product rule: Write it as \( 2 \cdot e^t \cdot t^{1/2} \). Using the product rule: \[ \frac{d}{dt} (e^t \cdot t^{1/2}) = e^t \cdot \frac{1}{2} t^{-1/2} + t^{1/2} \cdot e^t \] So, \[ \frac{d}{dt} (2e^t \sqrt{t}) = 2 \left( e^t \cdot \frac{1}{2} t^{-1/2} + t^{1/2} \cdot e^t \right) = e^t t^{-1/2} + 2t^{1/2} e^t \] Combine the derivatives: \[ g’(t) = 3t^2 - 4t + e^t t^{-1/2} + 2t^{1/2} e^t \] Now, find \( g’(1) \): \[ g’(1) = 3(1)^2 - 4(1) + e^1 \cdot 1^{-1/2} + 2 \cdot 1^{1/2} \cdot e^1 \] \[ = 3 - 4 + e + 2e \] \[ = -1 + 3e \] Next, find \( f’(t) \). Given \( f(t) = \frac{g(t)}{t^2} \). Use the quotient rule: \[ f’(t) = \frac{t^2 \cdot g’(t) - g(t) \cdot 2t}{(t^2)^2} \] Simplify: \[ f’(t) = \frac{t^2 g’(t) - 2tg(t)}{t^4} \] \[ = \frac{t g’(t) - 2g(t)}{t^3} \] Now, find \( f’(1) \): Evaluate \( g(1) \): \[ g(1) = 1^3 - 2(1)^2 + 2e^1 \cdot 1^{1/2} \] \[ = 1 - 2 + 2e \] \[ = -1 + 2e \] Substitute in \( f’(1) \): \[ f’(1) = \frac{1(-1 + 3e) - 2(-1 + 2e)}{1^3} \] \[ = (-1 + 3e + 2 - 4e) \] \[ = 1 - e \] Therefore, \( g’(1) = -1 + 3e \) \( f’(1) = 1 - e \)