The function f(x) = (x - 3)^2 satisfies all t | vedclass

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

Question

(B) According to the Mean Value Theorem,there exists a point $c \in (3, 4)$ such that $f'(c) = \frac{f(4) - f(3)}{4 - 3}$.First,calculate the slope of

Vedclass · Accepted Answer

$\left( \frac{7}{2}, \frac{1}{4} \right)$

Vedclass · Answer

(B) According to the Mean Value Theorem,there exists a point $c \in (3, 4)$ such that $f'(c) = \frac{f(4) - f(3)}{4 - 3}$.First,calculate the slope of the chord joining $(3, 0)$ and $(4, 1)$:$m = \frac{1 - 0}{4 - 3} = 1$.Now,find the derivative of $f(x) = (x - 3)^2$:$f'(x) = 2(x - 3)$.Set the derivative equal to the slope of the chord:$2(c - 3) = 1 \Rightarrow c - 3 = \frac{1}{2} \Rightarrow c = \frac{7}{2}$.Now,find the corresponding $y$-coordinate by substituting $x = \frac{7}{2}$ into the function:$y = \left( \frac{7}{2} - 3 \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.Thus,the point is $\left( \frac{7}{2}, \frac{1}{4} \right)$.

The function $f(x) = (x - 3)^2$ satisfies all the conditions of the Mean Value Theorem in $[3, 4]$. $A$ point on $y = (x - 3)^2$,where the tangent is parallel to the chord joining $(3, 0)$ and $(4, 1)$,is

Similar Questions

The function $f(x) = x^3 - 6x^2 + ax + b$ satisfies the conditions of Rolle's theorem in $[1, 3]$. Then the values of $a$ and $b$ are respectively

If Rolle's theorem holds for the function $f(x) = 2x^3 + ax^2 + bx$ in the interval $[-1, 1]$ for the point $c = \frac{1}{2}$,then the value of $2a + b$ is

Verify Rolle's Theorem for the function $f(x)=x^{2}+2x-8, x \in[-4,2]$

$A$ function $f$ is defined by $f(x)=2+(x-1)^{2/3}$ on $[0,2]$. Which of the following statements is incorrect?

The value of $c$ in the Lagrange's mean value theorem for $f(x)=\sqrt{x-2}$ in the interval $[2,6]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module