If for f(x) = 2x - x^2,Lagrange's Mean Value | vedclass

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

Question

(B) According to Lagrange's Mean Value Theorem,there exists a point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = 2x - x^

Vedclass · Accepted Answer

$c = \frac{1}{2}$

Vedclass · Answer

(B) According to Lagrange's Mean Value Theorem,there exists a point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = 2x - x^2$ on the interval $[0, 1]$,where $a = 0$ and $b = 1$.First,calculate $f(a)$ and $f(b)$:$f(0) = 2(0) - (0)^2 = 0$$f(1) = 2(1) - (1)^2 = 1$Next,calculate the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(2x - x^2) = 2 - 2x$Now,substitute these values into the theorem formula:$f'(c) = \frac{f(1) - f(0)}{1 - 0}$$2 - 2c = \frac{1 - 0}{1}$$2 - 2c = 1$$2c = 1$$c = \frac{1}{2}$Thus,the value of $c$ is $\frac{1}{2}$.

If for $f(x) = 2x - x^2$,Lagrange's Mean Value Theorem satisfies in $[0, 1]$,then the value of $c \in [0, 1]$ is

Similar Questions

Let $f$ be differentiable for all $x$. If $f(1) = -2$ and $f'(x) \ge 2$ for $x \in [1, 6]$,then

Let $f: R \rightarrow R$ be a twice continuously differentiable function such that $f(0) = f(1) = f^{\prime}(0) = 0$. Then:

The constant $c$ of Rolle's theorem for the function $f(x)=(x-1)^3(x-2)^5$ in the interval $[1, 2]$ is:

The number of admissible values of $C$ obtained when the Lagrange's mean value theorem is applied for the function $f(x)=x$ on $[2,5]$ is

Let $f(1) = -2$ and $f'(x) \ge 4.2$ for $1 \le x \le 6$. The possible value of $f(6)$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module