Find the approximate value of (255)^{frac{1}{ | vedclass

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

Question

(A) Let $f(x) = x^{\frac{1}{4}}$.We know that $256^{\frac{1}{4}} = 4$.Let $x = 256$ and $\Delta x = -1$.Then,$f(x + \Delta x) = (x + \Delta x)^{\frac{

Vedclass · Accepted Answer

$3.9961$

Vedclass · Answer

(A) Let $f(x) = x^{\frac{1}{4}}$.We know that $256^{\frac{1}{4}} = 4$.Let $x = 256$ and $\Delta x = -1$.Then,$f(x + \Delta x) = (x + \Delta x)^{\frac{1}{4}} = (255)^{\frac{1}{4}}$.The approximate value is given by $f(x + \Delta x) \approx f(x) + f'(x) \Delta x$.Here,$f'(x) = \frac{1}{4} x^{-\frac{3}{4}} = \frac{1}{4 x^{\frac{3}{4}}}$.For $x = 256$,$f'(256) = \frac{1}{4 (256)^{\frac{3}{4}}} = \frac{1}{4 (4^4)^{\frac{3}{4}}} = \frac{1}{4 	imes 4^3} = \frac{1}{4 	imes 64} = \frac{1}{256} = 0.00390625$.Thus,$f(255) \approx f(256) + f'(256) \Delta x = 4 + (0.00390625)(-1) = 4 - 0.00390625 = 3.99609375$.Rounding to four decimal places,we get $3.9961$.

Find the approximate value of $(255)^{\frac{1}{4}}$.

Similar Questions

The approximate value of $\tan ^{-1}(0.999)$ (up to $4$ decimal places) is

If there is an error of $\pm 0.04 \text{ cm}$ in the measurement of the diameter of a sphere,then the approximate percentage error in its volume,when the radius is $10 \text{ cm}$,is

The semi-vertical angle of a right circular cone is $45^{\circ}$. If the radius of the base of the cone is measured as $14 \text{ cm}$ with an error of $\left(\frac{\sqrt{2}-1}{11}\right) \text{ cm}$,then the approximate error in measuring its total surface area is (in $\text{sq. cm}$):

The radius and the height of a right circular solid cone are measured as $7 \text{ ft}$ each. If there is an error of $0.002 \text{ ft}$ for every foot in measuring them,then the error in the total surface area of the cone (in $\text{sq. ft}$) is:

Find the approximate value of $(32.15)^{\frac{1}{5}}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module