2. Given the function \( f(x)=2 x^{2}+3 x+1 \), a. Find the instantaneous rate of change when \( x=1 \) using the secant method. [4 marks] b. Find the value of the derivative at \( x=1 \) using first principles. [ 4 marks]

To solve the problem, we will address both parts step by step. ### Part a: Instantaneous Rate of Change using the Secant Method The instantaneous rate of change of a function at a point can be approximated using the secant method by calculating the slope of the secant line between two points on the function. We can choose two points close to \( x = 1 \), say \( x = 1 \) and \( x = 1 + h \) for a small \( h \). 1. **Choose \( h \)**: Let's take \( h = 0.01 \). 2. **Calculate \( f(1) \)** and \( f(1 + h) \)**: \[ f(1) = 2(1)^2 + 3(1) + 1 = 2 + 3 + 1 = 6 \] \[ f(1 + 0.01) = f(1.01) = 2(1.01)^2 + 3(1.01) + 1 \] First, calculate \( (1.01)^2 \): \[ (1.01)^2 = 1.0201 \] Now substitute: \[ f(1.01) = 2(1.0201) + 3(1.01) + 1 = 2.0402 + 3.03 + 1 = 6.0702 \] 3. **Calculate the slope of the secant line**: \[ \text{slope} = \frac{f(1 + h) - f(1)}{h} = \frac{f(1.01) - f(1)}{0.01} = \frac{6.0702 - 6}{0.01} = \frac{0.0702}{0.01} = 7.02 \] Thus, the instantaneous rate of change of \( f(x) \) at \( x = 1 \) using the secant method is approximately **7.02**. ### Part b: Value of the Derivative at \( x=1 \) using First Principles The derivative of a function at a point can be defined using the limit of the difference quotient: \[ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \] For \( a = 1 \): \[ f'(1) = \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \] 1. **Calculate \( f(1 + h) \)**: \[ f(1 + h) = 2(1 + h)^2 + 3(1 + h) + 1 \] Expanding \( (1 + h)^2 \): \[ (1 + h)^2 = 1 + 2h + h^2 \] Thus, \[ f(1 + h) = 2(1 + 2h + h^2) + 3(1 + h) + 1 = 2 + 4h + 2h^2 + 3 + 3h + 1 = 6 + 7h + 2h^2 \] 2. **Now substitute into the difference quotient**: \[ f'(1) = \lim_{h \to 0} \frac{(6 + 7h + 2h^2) - 6}{h} = \lim_{h \to 0} \frac{7h + 2h^2}{h} \] Simplifying: \[ f'(1) = \lim_{h \to 0} (7 + 2h) = 7 \] Thus, the value of the derivative at \( x = 1 \) using first principles is **7**. ### Summary of Results - Part a: Instantaneous rate of change at \( x = 1 \) using the secant method is approximately **7.02**. - Part b: Value of the derivative at \( x = 1 \) using first principles is **7**.