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Question: The value of \(x\) for which the value of the function \(f\left( x \right)=-\dfrac{11}{3}{{x}^{2}}+1...

The value of xx for which the value of the function f(x)=−113x2+17x−4313f\left( x \right)=-\dfrac{11}{3}{{x}^{2}}+17x-\dfrac{43}{13} is maximum is
(A) x=−10211x=-\dfrac{102}{11}
(B) x=−5122x=-\dfrac{51}{22}
(C) x=5122x=\dfrac{51}{22}
(D) x=10211x=\dfrac{102}{11}

Explanation

Solution

For answering this question we need to find the value of xx for which the value of the function f(x)=−113x2+17x−4313f\left( x \right)=-\dfrac{11}{3}{{x}^{2}}+17x-\dfrac{43}{13} is maximum. From the basic concept we know that the maximum of a function occurs when its derivative is zero.

Complete step by step answer:
Now from the question we have the function f(x)=−113x2+17x−4313f\left( x \right)=-\dfrac{11}{3}{{x}^{2}}+17x-\dfrac{43}{13} .
The value of xx for which the value of f(x)f\left( x \right) is maximum, f′(x)f'\left( x \right) should be zero.
So we need to differentiate the function.
For differentiating the function we will use the formulae
ddxax−2=−2ax\dfrac{d}{dx}a{{x}^{-2}}=-2ax . After applying this formula in the equation we have we will get
f′(x)=−113(2)x+17{{f}^{'}}\left( x \right)=-\dfrac{11}{3}\left( 2 \right)x+17 .
By equating it to zero we will have,
f′(x)=−113(2)x+17=0{{f}^{'}}\left( x \right)=-\dfrac{11}{3}\left( 2 \right)x+17=0
By simplifying this equation we will have,
−11(2)x+17(3)=0-11\left( 2 \right)x+17\left( 3 \right)=0
After further simplifications we will have,
−22x+51=0 ⇒22x=51 ⇒x=5122 \begin{aligned} & -22x+51=0 \\\ & \Rightarrow 22x=51 \\\ & \Rightarrow x=\dfrac{51}{22} \\\ \end{aligned}.
For a value of xx the function f(x)f\left( x \right) will have maximum or minimum value only when its derivative is equal to zero that is mathematically given as f′(x)=0f'\left( x \right)=0 and for the minimum of the function its double derivative will be positive given as f′′(x)>0f''(x)>0 and for maximum value it will be negative given as f′′(x)<0f''\left( x \right)<0 .
Here now we have the value of xx for which the first derivative of f(x)f\left( x \right) is zero. We need the value of xx for which it is maximum and it occurs when the second derivative is less than zero. Let us verify this.
The second derivative of f(x)f\left( x \right) is given as f′′(x)f''\left( x \right). We have the first derivative as f′(x)=−1132x+17f'\left( x \right)=-\dfrac{11}{3}2x+17 . The derivative of this is given as f′′(x)=−223f''\left( x \right)=-\dfrac{22}{3} . For the value of xx, for which is negative it indicates that the value of f(x)f\left( x \right) will be maximum.
We had come to a conclusion after performing the simplifications and shifting the values from the Left hand side to right hand side.
The conclusion we have got from this is the value of xx for which the value of the function f(x)=−113x2+17x−4313f\left( x \right)=-\dfrac{11}{3}{{x}^{2}}+17x-\dfrac{43}{13} is maximum is given as x=5122x=\dfrac{51}{22} .

So, the correct answer is “Option C”.

Note: While answering this type of questions we need to be clear with calculations and we should remember a further point which states that for a value of xx the function f(x)f\left( x \right) will have maximum or minimum value only when its derivative is equal to zero that is mathematically given as f′(x)=0f'\left( x \right)=0 and for the maximum of the function its double derivative will be negative given as f′′(x)<0f''(x)<0 and for minimum value it will be positive given as f′′(x)>0f''\left( x \right)>0 .