If y = x^3 + 5 and x changes from 3 to 2.99,t | vedclass

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

Question

(B) Given the function $y = x^3 + 5$.We need to find the approximate change in $y$ $(dy)$ when $x$ changes from $x = 3$ to $x = 2.99$.Here,$x = 3$ and

Vedclass · Accepted Answer

$-0.27$

Vedclass · Answer

(B) Given the function $y = x^3 + 5$.We need to find the approximate change in $y$ $(dy)$ when $x$ changes from $x = 3$ to $x = 2.99$.Here,$x = 3$ and the change in $x$ is $dx = 2.99 - 3 = -0.01$.The derivative of $y$ with respect to $x$ is $\frac{dy}{dx} = 3x^2$.Therefore,the approximate change $dy$ is given by $dy = (3x^2) \cdot dx$.Substituting the values $x = 3$ and $dx = -0.01$:$dy = 3(3)^2 \cdot (-0.01)$$dy = 3(9) \cdot (-0.01)$$dy = 27 \cdot (-0.01)$$dy = -0.27$.Thus,the approximate change in $y$ is $-0.27$.

If $y = x^3 + 5$ and $x$ changes from $3$ to $2.99$,then the approximate change in $y$ is

Similar Questions

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

The approximate value of the function $f(x) = x^{3} + 5x^{2} - 7x + 10$ at $x = 1.1$ is (in $.6$)

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

The population of a city grows at the annual rate of $3 \%$. What percentage increase is expected in $5 \text{ yr}$ (in $\%$)?

The approximate change in the volume $V$ of a cube of side $x$ meters caused by increasing the side by $3\%$ is: (in $x^{3} \text{ m}^{3}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module