How do you use linear approximation to the square root function to estimate square roots sqrt 8.95√8.95?

1 Answer
Jim H
Dec 3, 2016

The linear approximation of ff at aa is L(x) = f(a) + f'(a)(x-a). This is one way of writing the equation of the line tangent to the graph of f at (a,f(a))

Explanation:

The function we want to approximate is f(x) = sqrtx.

We want to estimate f(8.95) = sqrt 8.95 so we need a value for a that is close to 8.95 and for which we can find f(a). We can readily find f(1) or f(4), but f(5) would be more challenging.

Although we can easily find f(25), we'd like a to be as close as we can get to 8.95 and still calculate f(a).

By now you may be anxious for me to say, "we'll use a=9". So I will. We'll use a=9.

So f(a) = f(9) = 3

For the linearization we'll need f'(9) as well.

f'(x) = 1/(2sqrtx), so f'(9) = 1/6

The linearization of f(x) = sqrtx at a=9 is

L(x) = 3+1/6(x-9)

Our estimate is

f(8.95) ~~ L(8.95) = 3+1/6(8.95-9)

= 3-0.05/6 = 3- 0.008bar(3)

= 2.991bar6

Round to the desired accuracy. Keep in mind that this is an estimate for the desired number sqrt8.95 so don't claim too much accuracy.

2.99 or perhaps even 2.991 seems reasonable. (Although 2.9917 is accurate rounded to 4 decimal places.)

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