Verify the given linear approximation at a = 0. Then determine the values of x for which the linear approximation is accurate to within 0.1. (Enter your answer using interval notation. Round your answers to three decimal places.) 4 1 + 2x ≈ 1 + 1 2 x

x ∈ [-0.369, 0.678]

Step-by-step explanation:

To prove - ≈ 1 + x

As given , f(x) =

As we know that the linear approximation can be defined as :

L(x) = f(a) + f'(a)(x-a)             .........(1)

Now,

We have to show that at a = 0 , equation (1) satisfies

Now,

f(0) = = 1

Also,

f'(x) =

∴ we get

f'(x) =

⇒f'(0) =

Now, equation (1) becomes

L(x) = f(0) + f'(0)(x-0)

     = 1 + (x)

⇒ ≈ 1 + x

Hence verified.

Now,

| - (1 + x ) | ≤ 0.1

⇒ -0.36893 ≤ x ≤ 0.67766

⇒x ∈ [-0.369, 0.678]

the answer is 0.5 mpm

step-by-step explanation:

the line goes half way through the square each time (which is 0.5 or half of course) so that would mean its 0.5 mpm.

step-by-step explanation:

equation

you are asked to write the equation

  expression = 0

using the given expression. when you do that, you have

  2x^2-2x-12 = 0

factoring

you can see that each of the coefficients is even, so a factor of 2 can be removed immediately.

  2(x^2 -x -6) = 0

if you multiply the binomials (x+a)(x+b), you get

  (x +a)(x +b) = x^2 +(a+b)x +ab

this tells you the coefficient of x is the sum of two factors of the constant term. here, that means -1 is the sum of two factors of -6. we want to know those factors so we can factor the trinomial.

  -6 = -1×6 = -2×3 = -3×2 = -6×1

the factor pair that sums to -1 is {-3, 2}, so our factorization is

  2(x -3)(x +2) = 0

solutions

the solutions to this equation are the values of x that make the product zero. a product is zero only when only of the factors of the product is zero. here, that means

  x - 3 = 0

  x = 3 . . . add 3 to both sides of the equation

or

  x +2 = 0

  x = -2 . . . add -2 to both sides of the equation

the solutions to the equation are x=3 or x=-2. both of these values of x satisfy the equation—they make it be true.