Let f(x) be a continuous and differentiable f | vedclass

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

Question

(C) By the definition of the derivative,$f'(x) = \lim_{h 	o 0} \frac{f(x + h) - f(x)}{h}$.Given the functional equation $f(x + y) = f(x) - 3xy + f(y)

Vedclass · Accepted Answer

$-3x + 7$

Vedclass · Answer

(C) By the definition of the derivative,$f'(x) = \lim_{h 	o 0} \frac{f(x + h) - f(x)}{h}$.Given the functional equation $f(x + y) = f(x) - 3xy + f(y)$,we substitute $y = h$:$f(x + h) = f(x) - 3xh + f(h)$.Substituting this into the derivative definition:$f'(x) = \lim_{h 	o 0} \frac{f(x) - 3xh + f(h) - f(x)}{h}$$f'(x) = \lim_{h 	o 0} \frac{-3xh + f(h)}{h}$$f'(x) = \lim_{h 	o 0} (-3x + \frac{f(h)}{h})$.Given that $\lim_{h 	o 0} \frac{f(h)}{h} = 7$,we have:$f'(x) = -3x + 7$.

Let $f(x)$ be a continuous and differentiable function for all real numbers. If $f(x + y) = f(x) - 3xy + f(y)$ and $\lim_{h \to 0} \frac{f(h)}{h} = 7$,then the value of $f'(x)$ is:

Similar Questions

Let $g(x)$ be the anti-derivative of $f(x)$. Then the function for which $\log _e(1+(g(x))^2)+c$ is an anti-derivative is:

If $y = \frac{\sqrt{x^2 + 1} + \sqrt{x^2 - 1}}{\sqrt{x^2 + 1} - \sqrt{x^2 - 1}}$,then $\frac{dy}{dx} = $

Differentiate the function with respect to $x$: $\cos(x^{3}) \cdot \sin^{2}(x^{5})$

Let $f(x + y) = f(x)f(y)$ and $f(x) = 1 + \sin(3x)g(x)$,where $g(x)$ is continuous. Then $f'(x)$ is:

If $G(x) = -\sqrt{25 - x^2}$,then $\mathop{\lim}\limits_{x \to 1} \frac{G(x) - G(1)}{x - 1} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module