If f(x)=x2−25x−5f(x)=x2−25x−5, then is it possible to find f′(5)f′(5) since x≠5x≠5for the original function?

Answer 1

f'(x) = 2x-25, f'(5) = 2(5)-25 = 10-25 =-15

-15 is the slope of a tangent line of f(x) at x=5