The approximate value of f(x)=x^3+5x^2-7x+9 a | vedclass

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

Question

(A) Using the linear approximation formula: $f(x+h) \approx f(x) + h f'(x)$.Here,let $x=1$ and $h=0.1$.First,calculate $f(1) = (1)^3 + 5(1)^2 - 7(1) +

Vedclass · Accepted Answer

$8.6$

Vedclass · Answer

(A) Using the linear approximation formula: $f(x+h) \approx f(x) + h f'(x)$.Here,let $x=1$ and $h=0.1$.First,calculate $f(1) = (1)^3 + 5(1)^2 - 7(1) + 9 = 1 + 5 - 7 + 9 = 8$.Next,find the derivative $f'(x) = 3x^2 + 10x - 7$.Calculate $f'(1) = 3(1)^2 + 10(1) - 7 = 3 + 10 - 7 = 6$.Now,substitute these values into the formula:$f(1.1) \approx f(1) + 0.1 	imes f'(1)$$f(1.1) \approx 8 + 0.1 	imes 6$$f(1.1) \approx 8 + 0.6 = 8.6$.

The approximate value of $f(x)=x^3+5x^2-7x+9$ at $x=1.1$ is

Similar Questions

The approximate value of $\cos(30^{\circ} 30^{\prime})$ is,given that $1^{\circ} = 0.0175 \text{ rad}$ and $\cos 30^{\circ} = 0.8660$.

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

Find the approximate value of $(1.999)^{5}$.

Find the approximate value of $(82)^{\frac{1}{4}}$.

The circumference of a circle is measured as $56 \text{ cm}$ with an error of $0.02 \text{ cm}$. The percentage error in its area is

Vedclass Test Series

Exam Paper Generator

Online Exam Module