evaluate f(3)
evaluate (3+h) and simplify
then take
Limit [ (f(3+h) - f(3) ]/h as h --->0
You should be able to factor out an h from the top which would then cancel the h in the bottom
you should get
f'(3) = 138
b) when x=3, y = 138 (that's a coincidence)
so you have
slope of tangent is 138 and (3,138) is a point on it
y = 138x + b , plug in the point
138 = 138(3) + b
b = -276
so y = 138x - 276 is the tangent equation
Consider the function.
f (x) = 5x3 + 3x
(a) Compute the derivative at x = 3 using the limit definition.
f '(3) =
(b) Find an equation of the tangent line.
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