3. Let \( f(x)=a x^{3}+x-10=0 \). If the Newton method is used starting with the initial value \( x_{0}=1 \) then the first iteration value is found as \( x_{1}=2 \) then what is the value of \( a \) ? \( \begin{array}{llll}\text { a. } 0 & \text { b. } 1 & \text { c. } 2 & \text { d. } 3\end{array} \)

To find the value of \(a\) using the information given, we first set up the Newton's method formula: \[ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \] Given \(f(x) = ax^3 + x - 10\), we compute \(f'(x)\): \[ f'(x) = 3ax^2 + 1 \] Now, substituting our initial value \(x_0 = 1\) into \(f(x)\): \[ f(1) = a(1)^3 + 1 - 10 = a - 9 \] Next, we compute \(f'(1)\): \[ f'(1) = 3a(1)^2 + 1 = 3a + 1 \] Applying the Newton's method: \[ x_1 = 1 - \frac{f(1)}{f'(1)} = 1 - \frac{a - 9}{3a + 1} \] We know \(x_1 = 2\), so we set up the equation: \[ 2 = 1 - \frac{a - 9}{3a + 1} \] Rearranging gives us: \[ 1 = -\frac{a - 9}{3a + 1} \] Multiplying both sides by \(3a + 1\): \[ 3a + 1 = - (a - 9) \] Expanding and collecting terms: \[ 3a + 1 = -a + 9 \implies 4a = 8 \implies a = 2 \] Thus, the value of \(a\) is **2**. Therefore, the answer is: **c. 2**