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

I have tried entering 2.98328 but that does not appear to be correct.

Google search says it's 2.98329

WebWorks will not take the answer.

Did you look at my previous answer to this problem?

Why are you changing names?

http://www.jiskha.com/display.cgi?id=1412554250

It's 2.983, you only need 4 decimal places

To find the linear approximation of the square root of a number, we can use a process called linearization. The formula for linearization is given by:

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

Where:
L(x) is the linear approximation of the function f(x),
f(a) is the value of the function at a,
f'(a) is the derivative of the function at a,
x is the input to the function,
a is the point at which we are linearizing the function.

In this case, we want to find the linear approximation of the square root of 8.9 accurate to the fifth decimal place. Let's use a = 9 as the point for linearization. The function we are linearizing is f(x) = √x.

First, let's calculate f(a) and f'(a):
f(a) = √(9) = 3
f'(a) = (1 / (2√a)) = (1 / (2√9)) = 1/6

Now, substitute these values into the linearization formula:
L(x) = f(a) + f'(a) * (x - a)
L(x) = 3 + (1/6) * (x - 9)

To find the linear approximation of the square root of 8.9, substitute x = 8.9 into the formula:
L(8.9) = 3 + (1/6) * (8.9 - 9)
L(8.9) = 3 + (1/6) * (-0.1)
L(8.9) = 3 - 0.01667
L(8.9) = 2.98333 (rounded to decimal place)

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