Find all relative extrema of the function. Use the second derlvative test where applicable. (If an answer does not exist, enter DNE.) \[ f(x)=x^{2}+7 x-2 \] relative maximum \( \quad(x, y)=(\sqrt{\text { DNE }} \) relative minimum ( \( x, y)=(\sqrt{\text { DN }} \) )

To find the relative extrema of the function \( f(x) = x^2 + 7x - 2 \), we first need to compute the first derivative: \[ f'(x) = 2x + 7. \] Setting \( f'(x) = 0 \) to find critical points: \[ 2x + 7 = 0 \] \[ 2x = -7 \] \[ x = -\frac{7}{2} \] Next, we need the second derivative: \[ f''(x) = 2. \] Since \( f''(x) = 2 \) is positive, it indicates that the critical point \( x = -\frac{7}{2} \) is a relative minimum. Now, we can find the corresponding \( y \)-value by plugging \( x = -\frac{7}{2} \) into the original function: \[ f\left(-\frac{7}{2}\right) = \left(-\frac{7}{2}\right)^2 + 7\left(-\frac{7}{2}\right) - 2 \] \[ = \frac{49}{4} - \frac{49}{2} - 2 \] \[ = \frac{49}{4} - \frac{98}{4} - \frac{8}{4} \] \[ = \frac{49 - 98 - 8}{4} = \frac{-57}{4}. \] Now we can provide the final answers: Relative maximum: \( (x, y) = (\sqrt{\text{DNE}}) \) Relative minimum: \( (x, y) = \left(-\frac{7}{2}, -\frac{57}{4}\right) \)