Video example example 1 find the linearization of the function f(x) = x + 1 at a = 3 and use it to approximate the numbers 3.98 and 4.05 . are these approximations overestimates or underestimates? solution the derivative of f(x) = (x + 1)1/2 is f '(x) = correct: your answer is correct. and so we have f(3) = 2 correct: your answer is correct. and f '(3) = 1/4 correct: your answer is correct. putting these values into this equation, we see that the linearization is

  the linearization is y = 1/4x +5/4

  the linearization will produce overestimates

  the values computed from this linearization are ...

     f(3.98) ≈ 2.245

     f(4.05) ≈ 2.2625

Step-by-step explanation:

Apparently, you have ...

from which you have correctly determined that ...

so that f(3) = 2 and f'(3) = 1/4. Putting these values into the point-slope form of the equation of a line, we get the linearization ...

  g(x) = (1/4)(x -3) +2

  g(x) = (1/4)x +5/4

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The values from this linearization will be overestimates, as the curve f(x) is concave downward everywhere. The tangent (linearization) is necessarily above the curve everywhere.

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At the given values, we find ...

  g(3.98) = 2.245

  g(4.05) = 2.2625

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step-by-step explanation:

. + λrh() e 1 1 2 4 4 4 4 4 3 4 4 4 4 4 + th−1(− s(x)2 dx ω + λrh−1() e 2 1 2 4 4

step-by-step explanation: