If a function f is differentiable on R such t | vedclass

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

Question

(B) By the Mean Value Theorem,for any interval $[a, b]$,there exists $c \in (a, b)$ such that $f^{\prime}(c) = \frac{f(b) - f(a)}{b - a}$.Applying thi

Vedclass · Accepted Answer

$[0, 2]$

Vedclass · Answer

(B) By the Mean Value Theorem,for any interval $[a, b]$,there exists $c \in (a, b)$ such that $f^{\prime}(c) = \frac{f(b) - f(a)}{b - a}$.Applying this to the interval $[2, 4]$,we have $f^{\prime}(c_1) = \frac{f(4) - f(2)}{4 - 2} = \frac{f(4) + 6}{2}$.Since $f^{\prime}(x) \leq 4$,we have $\frac{f(4) + 6}{2} \leq 4$,which implies $f(4) + 6 \leq 8$,so $f(4) \leq 2$.Applying the Mean Value Theorem to the interval $[4, 6]$,we have $f^{\prime}(c_2) = \frac{f(6) - f(4)}{6 - 4} = \frac{8 - f(4)}{2}$.Since $f^{\prime}(x) \leq 4$,we have $\frac{8 - f(4)}{2} \leq 4$,which implies $8 - f(4) \leq 8$,so $f(4) \geq 0$.Combining these,we get $0 \leq f(4) \leq 2$,which means $f(4) \in [0, 2]$.

If a function $f$ is differentiable on $R$ such that $f^{\prime}(x) \leq 4$ for all $x \in R$; and if $f(2)=-6$ and $f(6)=8$,then the value of $f(4)$ belongs to the interval

Similar Questions

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

Let $f$ and $g$ be differentiable on the interval $I$ and let $a, b \in I, a < b$. Then,

Let $f:[a, b] \rightarrow R$ be such that $f$ is differentiable in $(a, b)$, continuous at $x=a$ and $x=b$, and $f(a)=0=f(b)$. Then:

Let $f:[1,3] \rightarrow R$ be a continuous function that is differentiable in $(1,3)$ and $f^{\prime}(x)=|f(x)|^{2}+4$ for all $x \in(1,3).$ Then,

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module