Question

Find the maximum value of the fraction \dfrac{17}{3}-{{\left( x-\dfrac{4}{5} \right)}^{2}}$$$$$ A. \dfrac{4}{5} B. $\dfrac{-4}{5}
C. \dfrac{17}{3}$$$$$ D. \dfrac{-17}{3}$$$$$

Answer 1

Use the concept of first and second order derivative test to find out the maximum value of the given expression.

Complete step by step answer:
Maxima and minima of a function :-
We know from differential calculus us that if $f\left( x \right)$ is a real valued single function defined within some interval $I$ , it will have its global maximum value (called maxima ) $x=a$ when $f\left( a \right) < f\left( x \right)$ for all $x\in I$ and global minimum value $x=a$ (called minima ) when $f\left( a \right)>f\left( x \right)$ for all $x\in I$. The function $f\left( x \right)$ will have its local maxima or minima at a point $x=a$ when $f\left( a \right)$ is more or less respectively at the neighbourhood points. Mathematically, $f\left( x \right)$ have a local maxima at $x=a$ and for very small positive quantity $h$ if we find $f\left( a \right)>f\left( a-h \right)$ and $f\left( a \right)>f\left( a+h \right)$$$$$ Critical Points:- The critical points of a function are the points where first order derivative of the function zero does not exist. Mathematically, $x=c$ is a critical point if $f'(c)=0$ or $f'(c)$ does not exist. The minima or maxima occur only at critical points.
The first order derivative test:- Let If $x=c$ is a critical point and 1\. $f\left( x \right)$ has a local maxima when ${{f}^{'}}\left( x \right)$ changes sign from positive to negative as we increase though the point $x=c$.
2. $f\left( x \right)$ has a local minima when ${{f}^{'}}\left( x \right)$ changes sign from negative to positive as we increase though the point x=c$$$$$ 3\. f\left( x \right)$will not have maxima or minima${{f}^{'}}\left( c \right)=0$$$$ The second order derivative test-$$$$ Iff(x)$is two times differentiable function,$x=c$is a critical point and${{f}^{'}}\left( c \right)=0 1\. $f\left( x \right)$ has a local maxima at $x=c$ if ${{f}^{''}}\left( c \right)<0
2. $f\left( x \right)$ has a local minima at $x=c$ if {{f}^{''}}\left( c \right)>0$$$$$ 3\. If both {{f}^{'}}\left( c \right)=0$and${{f}^{''}}\left( c \right)=0, then we move to third order derivative test.$$$$ Let us denote the given expression as f\left( x \right)$and find out its first and second derivative.$$$$

& f\left( x \right)=\dfrac{17}{3}-{{\left( x-\dfrac{4}{5} \right)}^{2}} \\\ & {{f}^{'}}\left( x \right)=-2\left( x-\dfrac{4}{5} \right) \\\ & {{f}^{''}}\left( x \right)=-2x \\\ \end{aligned}$$ We determine the critical points by assigning ${{f}^{'}}\left( x \right)=0$. $$\begin{aligned} & {{f}^{'}}\left( x \right)=0 \\\ & \Rightarrow -2\left( x-\dfrac{4}{5} \right)=0 \\\ & \Rightarrow x=\dfrac{4}{5} \\\ \end{aligned}$$ We find a critical point $x=\dfrac{4}{5}$ and see that ${{f}^{'}}\left( \dfrac{4}{5} \right)=0$.So we cannot find the maxima with first order derivative test and move onto second order derivative test. Now at $x=\dfrac{4}{5}$ $${{f}^{''}}\left( \dfrac{4}{5} \right)=-2\times \dfrac{4}{5}<0$$ So $f\left( x \right)$ will have local maxima at $x=\dfrac{4}{5}$ and the maximum value is $f\left( \dfrac{4}{5} \right)=\dfrac{17}{3}-{{\left( \dfrac{4}{5}-\dfrac{4}{5} \right)}^{2}}=\dfrac{17}{3}$ **So, the correct answer is “Option C”.** **Note:** The question tests your knowledge of maxima and minima of a function as an application derivative. Careful substitution and use of tests will lead us to the correct result. The question can also be framed to find out the minima. If you know coordinate geometry and you will know that the expression is an expression of downward parabola.