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Find the approximate value of (1.999)^5.

Answers (1)

Given:  (1.999)^5
And as the nearest integer to 1.999 is 2 , \mathrm{So}, 1.999=2-0.001
\therefore a=2$ and $h=-0.001
Hence, (1.999)^{5}=(2+(-0.001))^{5}
Therefore, the function becomes, f(x)=x^{5} \ldots \ldots \ldots(i)
After applying the first derivative, we get f^{\prime}(x)=5 x^{4} \ldots \ldots \ldots . (ii)
Now let f(a+h)=(1.999)^{5}

Now we know,

f(a+h)=f(a)+h f^{\prime}(a)
From equations (i) and (ii), substituting of functions results in,

f(a+h)=a^{5}+h\left(5 a^{4}\right)
Substitution of values of a and h, we get

\\\Rightarrow f(1.999)=32+(-0.001)(5(16)) \\\Rightarrow(1.999)^{5}=32+(-0.001)(80)

\\ \Rightarrow(1.999)^{5}=32-0.08 \\ \Rightarrow(1.999)^{5}=31.92

So, the approximate value of (1.999)^5 = 31.92.

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