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.