Question

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

Answer 1

Answer:

  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 ...

  $f(x)=\sqrt{x+1}$

from which you have correctly determined that ...

  $f'(x)=\dfrac{1}{2\sqrt{x+1}}$

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

__

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.

__

At the given values, we find ...

  g(3.98) = 2.245

  g(4.05) = 2.2625