Answer:
[tex]f'(x) = -28 \ at \ x = 2[/tex]
Step-by-step explanation:
[tex]f'(2) = \lim_{h \to 0}\frac{f(2+h)) -f(2)}{h} \\\\f(2+h) = 1 - 7(2+h)^2 = 1 - 7(4 +h^2 +4h) = 1 - 28 - 7h^2 - 28h = - 7h^2 -28h - 27\\\\f(2) = 1 - 7(2^2) = 1 - 28 = -27\\\\f(2+h) - f(2) = -7h^2 - 28h - 27 - (-27) = -7h^2 -28h -27 + 27 = -7h^2 -28h\\\\f'(2) = \lim_{h \to 0}\frac{-7h^2 - 28h}{h} \\\\[/tex]
[tex]= \lim_{h \to 0} \frac{h(-7h -28)}{h}\\\\= \lim_{h \to 0} (-7h -28)}\\\\= -7 (0) - 28\\\\= -28[/tex]