(a)
(f(3+.03)-f(3))/.03 = (3.1209-3)/.03 = 4.03
(b)
f'(x) = 2x-2
f'(3) = 4
(c)
f(3) = 3
So, now you have a point and a slope. The line is easy to express.
(d)
Now just plug 2.98 into that line.
(e)
where is f'(x) = 0?
Let f(x) = x^2 -2x
a) Estimate f'(3) using h = 0.03
b) Find the exact value for f'(3)
c) Give the equation of the tangent line to f(x) at x=3
d) Use the tangent line to estimate f(2.98)
e) Critical points
2 answers
Thanks!