Given f(x) = x^3 - 4x,if x changes from 2 to | vedclass

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

Question

(B) Let $y = f(x) = x^3 - 4x$. We are given $x = 2$ and the change in $x$,denoted by $\Delta x = 1.99 - 2 = -0.01$. The approximate change in $y$,deno

Vedclass · Accepted Answer

$-0.08$

Vedclass · Answer

(B) Let $y = f(x) = x^3 - 4x$. We are given $x = 2$ and the change in $x$,denoted by $\Delta x = 1.99 - 2 = -0.01$. The approximate change in $y$,denoted by $\Delta y$,is given by $\Delta y \approx \frac{dy}{dx} \Delta x$. First,find the derivative of $f(x)$ with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(x^3 - 4x) = 3x^2 - 4$. Now,evaluate the derivative at $x = 2$: $\left. \frac{dy}{dx} ight|_{x=2} = 3(2)^2 - 4 = 3(4) - 4 = 12 - 4 = 8$. Finally,calculate the approximate change $\Delta y$: $\Delta y \approx 8 	imes (-0.01) = -0.08$.

Given $f(x) = x^3 - 4x$,if $x$ changes from $2$ to $1.99$,then the approximate change in the value of $f(x)$ is

Similar Questions

The approximate value of $(8.01)^{4/3} + (8.01)^2$ up to $3$ decimal places is

If an error of $0.02 \text{ cm}^2$ is found in the surface area of a sphere when its radius is measured as $10 \text{ cm}$, then the approximate error that occurs in the volume of the sphere, in cubic centimetres, is

The approximate value of $\sqrt{6560}$ is

Find the approximate value of $f(3.02),$ where $f(x)=3x^{2}+5x+3$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module