The value c of Lagrange's mean value theorem | vedclass

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

Question

(A) Given the function $f(x) = e^x + 24$ on the interval $[0, 1]$.According to Lagrange's Mean Value Theorem,there exists at least one point $c \in (0

Vedclass · Accepted Answer

$\log (e-1)$

Vedclass · Answer

(A) Given the function $f(x) = e^x + 24$ on the interval $[0, 1]$.According to Lagrange's Mean Value Theorem,there exists at least one point $c \in (0, 1)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Here,$a = 0$ and $b = 1$.First,calculate $f(0) = e^0 + 24 = 1 + 24 = 25$.Next,calculate $f(1) = e^1 + 24 = e + 24$.The derivative is $f'(x) = e^x$,so $f'(c) = e^c$.Substituting these into the formula: $e^c = \frac{(e + 24) - 25}{1 - 0}$.$e^c = e - 1$.Taking the natural logarithm on both sides,we get $c = \log(e - 1)$.

The value $c$ of Lagrange's mean value theorem for $f(x)=e^{x}+24$ in $[0,1]$ is

Similar Questions

If $a, b, c$ are real numbers such that $a + b + c = 0$,then the quadratic equation $3ax^2 + 2bx + c = 0$ has

Let $f(x)=x^3+2x^2-x$ be a real-valued function. Then, the value of Lagrange's constant $C$ in $(-1,2)$ is

Let the function $f:[-7,0] \rightarrow R$ be continuous on $[-7,0]$ and differentiable on $(-7,0)$. If $f(-7)=-3$ and $f'(x) \leq 2$ for all $x \in (-7,0)$,then for all such functions $f$,$f(-1)+f(0)$ lies in the interval:

Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then

Which of the following functions can satisfy Rolle's theorem on the given interval?

Vedclass Test Series

Exam Paper Generator

Online Exam Module