Find the approximate value of f(2.01), where | vedclass

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

Question

(A) Let $x=2$ and $\Delta x=0.01.$ Then,we have:$f(x+\Delta x) \approx f(x) + f'(x) \Delta x$Given $f(x) = 4x^{2} + 5x + 2$,we find the derivative:$f'

Vedclass · Accepted Answer

$28.21$

Vedclass · Answer

(A) Let $x=2$ and $\Delta x=0.01.$ Then,we have:$f(x+\Delta x) \approx f(x) + f'(x) \Delta x$Given $f(x) = 4x^{2} + 5x + 2$,we find the derivative:$f'(x) = 8x + 5$At $x=2$,$f(2) = 4(2)^{2} + 5(2) + 2 = 16 + 10 + 2 = 28$.At $x=2$,$f'(2) = 8(2) + 5 = 16 + 5 = 21$.Now,substitute these values into the approximation formula:$f(2.01) \approx f(2) + f'(2) \Delta x$$f(2.01) \approx 28 + (21)(0.01)$$f(2.01) \approx 28 + 0.21$$f(2.01) \approx 28.21$Thus,the approximate value of $f(2.01)$ is $28.21$.

Find the approximate value of $f(2.01),$ where $f(x)=4x^{2}+5x+2$.

Similar Questions

The approximate value of $\sqrt[3]{28}$ rounded up to $3$ decimal places is

If $y = (1 + \alpha + \alpha^2 + \ldots) e^{nx}$,where $\alpha$ and $n$ are constants,then the relative error in $y$ is

If the area of a square is $575$ square units,then the approximate value of its side is

Find the approximate value of $\sqrt{49.5}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module