# Find the approximate value of (1.999)5.

Given (1.999)5

But the integer nearest to 1.999 is 2,

So, 1.999 = 2-0.001

, a = 2 and h = -0.001

Hence, (1.999)5 = (2+(-0.001))5

Let the function becomes,

f(x) = x5………(i)

Now applying first derivative, we get

f’(x) = 5x4……….(ii)

Now let f(a+h) = (1.999)5

Now we know,

f(a+h) = f(a)+hf’(a)

Now substituting the function from (i) and (ii), we get

f(a+h) = a5+h(5a4)

Substituting the values of a and h, we get

f(2+(-0.001)) = 25+( -0.001) (5(24))

f(1.999) = 32+(-0.001)(5(16))

(1.999)5 = 32+(-0.001)(80)

(1.999)5 = 32-0.08

(1.999)5 = 31.92

Hence the approximate value of (1.999)5 = 31.92.

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