Posted by **Rachael** on Sunday, June 9, 2013 at 1:12am.

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**, Sunday, June 9, 2013 at 8:49am
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

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