9) Find \( f^{\prime}(x) \) using the limit desinition of the derivative. Then evaluate at \( x=-8 \). \[ f(x)=-x^{2}+4 x+4 \]

To find f '(x) using the limit definition, recall that   f '(x) = lim₍ₕ→0₎ [f(x + h) − f(x)] / h. Given f(x) = –x² + 4x + 4, first compute f(x + h):   f(x + h) = –(x + h)² + 4(x + h) + 4         = –(x² + 2xh + h²) + 4x + 4h + 4         = –x² – 2xh – h² + 4x + 4h + 4. Now, find the difference f(x + h) – f(x):   f(x + h) – f(x) = [ –x² – 2xh – h² + 4x + 4h + 4 ] – [ –x² + 4x + 4 ]             = –x² – 2xh – h² + 4x + 4h + 4 + x² – 4x – 4             = –2xh – h² + 4h. Factor h from the numerator:   = h(–2x – h + 4). The derivative becomes:   f '(x) = lim₍ₕ→0₎ [h(–2x – h + 4)] / h         = lim₍ₕ→0₎ (–2x – h + 4). Taking the limit as h approaches 0:   f '(x) = –2x + 4. Now, evaluate the derivative at x = –8:   f '(–8) = –2(–8) + 4         = 16 + 4         = 20. Thus, the derivative of f(x) is f '(x) = –2x + 4 and f '(–8) = 20.