Question

Find the approximate value of √8.95​

Answer 1

Step-by-step explanation:

Let f(x)=√x

Letf(x)=x

f(a+h)=√a+h

f(a+h)=a+h

we choose a=9 and h=-0.05

Then √8.95=f(a+h)

8.95=f(a+h)

But ,f(a+h)≈f(a)+hf'(a)

,f(a+h)≈f(a)+hf′(a)

√8.95≈f(a)+hf'(a)...........(1)

Now f(a)=√a=√9 and h=−0.05

f(a)=a=9andh=-0.05

we have

f(X)=√x

f(X)=x

f'(x)=12√x

f′(x)=12x

f'(a)=f'(9)=12√9=12×3

=16

f′(a)=f′(9)=129=12×3=16

from(1)

√8.95≈3+(−0.05)(16)

8.95≈3+(-0.05)(16)

=3-0.0083=2.9917

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Answer 2

Answer:2.9916550603

Step-by-step explanation:

Approximate value of√8.95 is 2.9916550603