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

__

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

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.