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(x-5)
y = 68x - 340 + 95
y = 68x - 245
If F(x)=x^3−7x+5, use the limit definition of the derivative to find F�Œ(5), then find an equation of the tangent line to the curve y=x^3−7x+5 at the point (5, 95).
F�Œ(5)=
The equation of the tangent line is y = x + .
Check your answer for yourself by graphing the curve and the tangent line.
1 answer