Calculus
posted by Rachael .
If F(x)=x^3−7x+5, use the limit definition of the derivative to find FŒ(5), then find an equation of the tangent line to the curve y=x^3−7x+5 at the point (5, 95).
FŒ(5)=
The equation of the tangent line is y = x + .
Check your answer for yourself by graphing the curve and the tangent line.

Calculus 
Reiny
f(5) = 125  35 + 5 = 95
f(5+h) = (5+h)^3  7(5+h) + 5
= h^3 + 15h^2 +68h + 95  (I'll let you check that
derivative = Lim ( f(h+5)  f(5) )/h as h > 0
= lim ( h^3 + 15 h^2 + 68h = 95  95)/h
= lim h^2 + 15h + 68 , as h > 0
= 68
so equation of tangent at (5,95)
y  95 = 68(x5)
y = 68x  340 + 95
y = 68x  245
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