What is the linear approximation of the square root of 8.9? Must be accurate to the fifth decimal place.

I have tried everything it seems like yet cannot get the right answer.

http://www.google.com/#q=square+root+of+8.9

If we let L(x) represent the linear approximation for some value of x

L(a) = f(a) + f'(a)/(x-a) , were a is a number for which I know the value of f(a)

so for √8.9
I know that √9 = 3 , so let a = 9

f(9) = √9 = 3
for f(x) = √x = x^(1/2)
f'(x) - (1/2x^(-1/2) = 1/(2√x)

so √8.9) = 3 + (1/6)(8.9-9)
= 3 - 1/60
I know 1/6 = .16666...
so 1/60 = .016666....
and √8.9 = appr 3 - .01666...
= appr 2.98333...

check with calculator: (which we are not supposed to have for this)
= 2.98329.. , not bad

Well, solving this math problem might not require a rubber nose and a red wig, but let's give it a shot! When it comes to approximating the square root of a number, you can use the linear approximation formula:

f(x) ≈ f(a) + f'(a)(x-a)

In this case, we want to approximate the square root of 8.9. Let's pick a number that's close to 8.9, like 9.

f(x) = √x, so f(a) = √9 = 3

The derivative of f(x) = √x is f'(x) = 1 / (2√x). So, f'(a) = 1 / (2√9) = 1 / 6

Now we can plug these values into the linear approximation formula:

f(x) ≈ f(a) + f'(a)(x-a)
≈ 3 + (1/6)*(8.9-9)
≈ 3 - 0.015
≈ 2.985

So, our approximation for the square root of 8.9 is approximately 2.985. Just remember, my accuracy might be clownish, so please verify!

To find the linear approximation of the square root of 8.9 accurate to the fifth decimal place, we can use the concept of Taylor series expansion.

The Taylor series expansion for a function f(x) centered at a point a is given by:

f(x) = f(a) + f'(a)(x - a) + f''(a)(x - a)^2/2! + f'''(a)(x - a)^3/3! + ...

In this case, we want to find the linear approximation by considering only the first two terms of the expansion:

f(x) ≈ f(a) + f'(a)(x - a)

Let's consider the function f(x) = √x and the point a = 9, which is closest to 8.9.

1. Find the value of f(a):

f(a) = √9 = 3

2. Find the derivative of f(x):

f'(x) = 1/(2√x)

3. Find the value of f'(a):

f'(a) = 1/(2√9) = 1/6

4. Substitute the values into the linear approximation formula:

f(x) ≈ f(a) + f'(a)(x - a)
√8.9 ≈ 3 + (1/6)(8.9 - 9)
√8.9 ≈ 3 - (1/6)(0.1)
√8.9 ≈ 3 - 0.0166667

Therefore, the linear approximation of the square root of 8.9 accurate to the fifth decimal place is approximately 2.98333.

To find the linear approximation of the square root of 8.9 accurate to the fifth decimal place, you can use the formula for a linear approximation:

L(x) = f(a) + f'(a)(x - a),

where L(x) is the linear approximation of the function f(x), a is the value of x around which you are approximating, and f'(a) is the derivative of the function evaluated at a.

In this case, f(x) = sqrt(x), and we are trying to approximate the square root of 8.9. Let's choose a value close to 8.9, such as a = 9. Note that f(x) = sqrt(x) is a continuous function, so the closer a is to 8.9, the more accurate the approximation will be.

First, we need to find f'(x) and evaluate it at a = 9. The derivative of f(x) = sqrt(x) can be found using calculus:

f'(x) = (1/2)sqrt(x).

Evaluating at a = 9, f'(9) = (1/2)sqrt(9) = (1/2) * 3 = 3/2.

Now that we have f'(9) = 3/2, we can use the linear approximation formula:

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

L(x) = sqrt(9) + (3/2)(x - 9).

Simplify further:

L(x) = 3 + (3/2)(x - 9).

Now, substitute x = 8.9 into the formula to get the linear approximation:

L(8.9) = 3 + (3/2)(8.9 - 9).

L(8.9) = 3 + (3/2)(-0.1).

L(8.9) = 3 - 0.15.

L(8.9) = 2.85.

Therefore, the linear approximation of the square root of 8.9 accurate to the fifth decimal place is 2.85.

If you're having trouble getting this result, make sure to double-check your calculations and be careful with rounding errors.