Let f(x) be a differentiable function for all | vedclass

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

Question

(A) By the definition of the derivative, $f^{\prime}(x) = \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$.Given the functional equation $f(x+y) = f(x)

Vedclass · Accepted Answer

$-3x+7$

Vedclass · Answer

(A) By the definition of the derivative, $f^{\prime}(x) = \lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$.Given the functional equation $f(x+y) = f(x) + f(y) - 3xy$, we substitute $y = h$ to get $f(x+h) = f(x) + f(h) - 3xh$.Substituting this into the derivative formula:$f^{\prime}(x) = \lim _{h \rightarrow 0} \frac{f(x) + f(h) - 3xh - f(x)}{h}$.Simplifying the expression:$f^{\prime}(x) = \lim _{h \rightarrow 0} \frac{f(h) - 3xh}{h} = \lim _{h \rightarrow 0} \left( \frac{f(h)}{h} - 3x \right)$.Given that $\lim _{h \rightarrow 0} \frac{f(h)}{h} = 7$, we have:$f^{\prime}(x) = 7 - 3x$.

Let $f(x)$ be a differentiable function for all $x \in R$ and $f(x+y)=f(x)+f(y)-3xy$. If $\lim _{h \rightarrow 0} \frac{f(h)}{h}=7$, then $f^{\prime}(x)=$

Similar Questions

$A$ possible positive value of '$a$',for which $f^{\prime}(x)=0$ has equal roots,is

$\frac{d}{d x}\left[a \tan ^{-1} x+b \log \left(\frac{x-1}{x+1}\right)\right]=\frac{1}{x^4-1}$
$\Rightarrow a-2 b$ is equal to

If $y=a \sin x+b \cos x$,then $y^2+\left(\frac{d y}{d x}\right)^2$ is a

If $y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \infty$,then $\frac{dy}{dx} = $

If $f(2)=4$ and $f^{\prime}(2)=1$,then $\lim _{x \rightarrow 2} \frac{x f(2)-2 f(x)}{x-2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module