Find a) any critical values and b) any relative extrema. \( f(x)=x^{2}+2 x+4 \) a) Select the correct choice below and, if necessary, fill in the answer box within your choice. A. The critical value(s) of the function is/are (Use a comma to separate answers as needed.) B. The function has no critical values.

Find the first order derivative with respect to \( x \) for \( x^2+2x+4 \). Evaluate the derivative by following steps: - step0: Evaluate the derivative: \(\frac{d}{dx}\left(x^{2}+2x+4\right)\) - step1: Use differentiation rules: \(\frac{d}{dx}\left(x^{2}\right)+\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(4\right)\) - step2: Calculate: \(2x+2+0\) - step3: Remove 0: \(2x+2\) Solve the equation \( 2x+2=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(2x+2=0\) - step1: Move the constant to the right side: \(2x=0-2\) - step2: Remove 0: \(2x=-2\) - step3: Divide both sides: \(\frac{2x}{2}=\frac{-2}{2}\) - step4: Divide the numbers: \(x=-1\) a) The critical value of the function is \( x = -1 \). b) To find the relative extrema, we need to evaluate the second derivative of the function. Let's calculate the second derivative. Evaluate the derivative by following steps: - step0: Evaluate the derivative: \(\frac{d}{dx}\left(2x+2\right)\) - step1: Use differentiation rules: \(\frac{d}{dx}\left(2x\right)+\frac{d}{dx}\left(2\right)\) - step2: Calculate: \(2+0\) - step3: Remove 0: \(2\) The second derivative of the function is \( 2 \). Since the second derivative is positive, the critical point at \( x = -1 \) is a relative minimum. Therefore, the function has a relative minimum at \( x = -1 \).