Question

For the function $ f\left( x \right) = {x^2} - 6x + 8 $ , $ 2 \leqslant x \leqslant 4 $ , then the value of x for which $ f'\left( x \right) $ vanishes is:
(A) $ \dfrac{9}{4} $
(B) $ \dfrac{5}{2} $
(C) $ 3 $
(D) $ \dfrac{7}{2} $

Answer 1

Hint: In the given question, we are provided with a function and we have to find the value of the variable x for which the derivative of the function vanishes. So, we first find the derivative of the function using the power rule of differentiation and then equate it to zero for finding the required value of the variable.

Complete step-by-step answer:
So, we have the function $ f\left( x \right) = {x^2} - 6x + 8 $
We differentiate the function with respect to x using the power rule of differentiation $ \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} $ . So, we have,
 $ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2} - 6x + 8} \right) $
 $ \Rightarrow f'\left( x \right) = 2x - 6 $
Now, we equate the derivative of the function to zero as we have to find the value of the variable x for which the derivative of the function vanishes. So, we have,
 $ \Rightarrow f'\left( x \right) = 2x - 6 = 0 $
Now, we use the method of transposition to shift the terms in the equation and find the value of the variable x.
So, we get,
 $ \Rightarrow 2x = 6 $
Dividing both the sides of the equation by two, we get,
 $ \Rightarrow x = 3 $
So, the derivative of the function $ f\left( x \right) = {x^2} - 6x + 8 $ for $ 2 \leqslant x \leqslant 4 $ vanishes for $ x = 3 $ .
Hence, option (C) is the correct answer to the given question.
So, the correct answer is “Option C”.

Note: If we add, subtract, multiply or divide by the same number on both sides of a given algebraic equation, then both sides will remain equal. Method of transposition involves doing the exact same mathematical thing on both sides of an equation with the aim of simplification in mind. This method can be used to solve various algebraic equations like the one given in question with ease. We must take care of the calculations while doing such questions.