Question

if y = f (x) = 1/x+2 , x =8 and delta x = 0.02 then find dy and delta y​

Answer 1

SOLUTION

GIVEN

$\displaystyle \sf{f(x) = \frac{1}{x + 2} }$

x = 8 and Δx = 0.02

TO DETERMINE

dy and Δy

EVALUATION

Here it is given that

$\displaystyle \sf{f(x) = \frac{1}{x + 2} }$

Then

$\displaystyle \sf{f'(x) = - \frac{1}{{(x + 2)}^{2} } }$

Now

Δy

= Increment of y

= f(x+Δx) - f(x)

= f(8+0.02) - f(8)

$\sf{ = f(8.02) - f(8)}$

$\displaystyle \sf{= \frac{1}{8.02 + 2} - \frac{1}{8 + 2} }$

$\displaystyle \sf{= \frac{1}{10.02} - \frac{1}{10} }$

$\sf{ = 0.0998 - 0.1}$

$\sf{ = - 0.0002}$

Again

dy

= f'(x) Δx

= f'(8) × 0.02

$\displaystyle \sf{= - \frac{1}{{(8 + 2 )}^{2} } \times 0.02}$

$\displaystyle \sf{= - \frac{1}{{(10 )}^{2} } \times 0.02}$

$\sf{ = - 0.01 \times 0.02}$

$\sf{ = - 0.0002}$

FINAL ANSWER

Δy = - 0.0002 and dy = - 0.0002

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Answer 2

Step-by-step explanation:

since it is doubt ful answer because of its different answers in other books