Using the definition, calculate the derivative of the function. Then find the values of the derivative as specified. \( f(x)=2+x^{2} ; f^{\prime}(-3), f^{\prime}(0), f^{\prime}(1) \)

**Step 1: Use the Definition of the Derivative** The definition of the derivative is \[ f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}. \] **Step 2: Write \( f(x+h) \) for \( f(x)=2+x^2 \)** Substitute \( x+h \) into the function: \[ f(x+h)=2+(x+h)^2 = 2 + x^2 + 2xh + h^2. \] **Step 3: Compute \( f(x+h)-f(x) \)** Subtract \( f(x) \) from \( f(x+h) \): \[ f(x+h)-f(x) = \left[ 2+x^2+2xh+h^2 \right] - \left[ 2+x^2 \right] = 2xh+h^2. \] **Step 4: Form the Difference Quotient** Divide by \( h \): \[ \frac{f(x+h)-f(x)}{h} = \frac{2xh+h^2}{h} = 2x + h. \] **Step 5: Take the Limit as \( h \to 0 \)** \[ f'(x)=\lim_{h \to 0}(2x+h)=2x. \] Thus, the derivative of the function is \[ f'(x)=2x. \] **Step 6: Evaluate the Derivative at the Specified Points** 1. For \( x=-3 \): \[ f'(-3) = 2(-3) = -6. \] 2. For \( x=0 \): \[ f'(0) = 2(0) = 0. \] 3. For \( x=1 \): \[ f'(1) = 2(1) = 2. \]