If \( f(x)=(x-9) e^{x} \), find \( f^{\prime}(x) \) and \( f^{\prime \prime}(x) \) \( f^{\prime}(x)=\square \) \( f^{\prime \prime}(x)=\square \)

To find \( f'(x) \) and \( f''(x) \) for the function \( f(x) = (x-9)e^x \), we can use the product rule. First, we'll find \( f'(x) \): Let \( u = x-9 \) and \( v = e^x \). Then, by the product rule: \[ f'(x) = u'v + uv' \] Calculating \( u' \) and \( v' \): \[ u' = 1 \quad \text{and} \quad v' = e^x \] Now substituting back into the product rule gives us: \[ f'(x) = (1)(e^x) + (x-9)(e^x) \] \[ = e^x + (x-9)e^x \] \[ = (x - 8)e^x \] Next, we'll find \( f''(x) \). Again, apply the product rule on \( f'(x) = (x - 8)e^x \): Let \( u = x - 8 \) and \( v = e^x \). \[ f''(x) = u'v + uv' \] This gives: \[ u' = 1 \] \[ v' = e^x \] Substituting back yields: \[ f''(x) = (1)(e^x) + (x-8)(e^x) \] \[ = e^x + (x-8)e^x \] \[ = (x - 7)e^x \] In conclusion, we have: \( f^{\prime}(x) = (x - 8)e^x \) \( f^{\prime \prime}(x) = (x - 7)e^x \)