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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

  • Calc - ,

    (f(3+.03)-f(3))/.03 = (3.1209-3)/.03 = 4.03

    f'(x) = 2x-2
    f'(3) = 4

    f(3) = 3
    So, now you have a point and a slope. The line is easy to express.

    Now just plug 2.98 into that line.

    where is f'(x) = 0?

  • Calc - ,


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