Find the approximate value of (0.999)^{frac{1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $f(x) = x^{\frac{1}{10}}$.We need to find the value of $f(0.999)$.Let $x = 1$ and $\Delta x = -0.001$,so that $x + \Delta x = 0.999$.Using the

Vedclass · Accepted Answer

$0.9999$

Vedclass · Answer

(A) Let $f(x) = x^{\frac{1}{10}}$.We need to find the value of $f(0.999)$.Let $x = 1$ and $\Delta x = -0.001$,so that $x + \Delta x = 0.999$.Using the formula for approximation,$f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.Here,$f(x) = x^{\frac{1}{10}}$,so $f(1) = 1^{\frac{1}{10}} = 1$.The derivative is $f'(x) = \frac{1}{10} x^{\frac{1}{10} - 1} = \frac{1}{10} x^{-\frac{9}{10}} = \frac{1}{10 x^{\frac{9}{10}}}$.At $x = 1$,$f'(1) = \frac{1}{10(1)^{\frac{9}{10}}} = \frac{1}{10} = 0.1$.Now,$f(0.999) \approx f(1) + f'(1) \Delta x$.$f(0.999) \approx 1 + (0.1)(-0.001) = 1 - 0.0001 = 0.9999$.

Find the approximate value of $(0.999)^{\frac{1}{10}}$.

Similar Questions

Let $y = f(x) = 2x^{2} - 3x + 2$. The differential of $y$ when $x$ changes from $2$ to $1.99$ is

Using differentials,find the approximate value of $(33)^{-\frac{1}{5}}$.

Find the approximate value of $(0.0037)^{\frac{1}{2}}$.

Find the approximate value of $\sqrt{0.6}$ using differentials.

The nearest approximate value of $\sqrt{2023}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module