For the function f(x) = x^2 - 6x + 8 where 2 | vedclass

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

Question

(C) Given the function $f(x) = x^2 - 6x + 8$. To find the value of $x$ for which $f'(x)$ vanishes,we first find the derivative of $f(x)$ with respect

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Given the function $f(x) = x^2 - 6x + 8$. To find the value of $x$ for which $f'(x)$ vanishes,we first find the derivative of $f(x)$ with respect to $x$: $f'(x) = \frac{d}{dx}(x^2 - 6x + 8) = 2x - 6$. Now,set the derivative equal to zero: $2x - 6 = 0$. Solving for $x$: $2x = 6 \Rightarrow x = 3$. Since $x = 3$ lies within the interval $[2, 4]$,the value of $x$ for which $f'(x)$ vanishes is $3$.

For the function $f(x) = x^2 - 6x + 8$ where $2 \le x \le 4$,the value of $x$ for which $f'(x)$ vanishes is:

Similar Questions

$\frac{d}{dx} \left( \sqrt{3} \sin \left(2x + \frac{\pi}{3} \right) + \cos \left(2x + \frac{\pi}{3} \right) \right) = $ . . . . . .

Let $f(x)$ be a polynomial of degree $3$ such that $f(3)=1$,$f'(3) = -1$,$f''(3) = 0$ and $f'''(3)=12$. Then the value of $f'(1)$ is

Find the derivative of the following function: $\frac{\sec x-1}{\sec x+1}$

Find the derivative: $\frac{d}{dx}[\cos((1 - x^2)^2)] = ?$

The differential coefficient of $f(\sin x)$ with respect to $x,$ where $f(x) = \log x,$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module