Chapter 7.6, Problem 4OE

To determine

Whether the function $f\left(x\right)=2x-x^2$ has a maximum or minimum value and then give the value of $\left(x\right)$ associated with it.

Answer to Problem 4OE

The function $f\left(x\right)=2x-x^2$ will have maximum value and $x=1$ is the associated value of $\left(x\right)$ which will give the maximum value of function equal to $1$ .

Explanation of Solution

Given information: The given function is,

  $f\left(x\right)=2x-x^2\mathrm{....}\left(1\right)$

Formula used: The function of the form $f\left(x\right)=a{x}^2+bx+c$ where $a\ne 0$ will have the maximum value if $a<0$ and minimum value when $a>0$ . Mathematically,

  $\left(\begin{array}{l}f\left(x\right)=a{x}^2+bx+c\text{ is maximum for }a<0\\ \text{ is minimum for }a>0\end{array}\right)\mathrm{....}\left(2\right)$

Calculation: Comparing the formula (2) with (1) gives, $a=-1$ which is less than zero. Therefore the given function $f\left(x\right)=2x-x^2$ will give the maximum value.

Now the value for $\left(x\right)$ associated with maximum value can be evaluated from the formula $x=-\frac{b}{2a}$ from (1),

  $\begin{array}{c}x=-\frac{\left(2\right)}{2\left(-1\right)}\mathrm{....}\left(a=-1,b=2\right)\\ x=1\end{array}$

Hence the value of function will give the maximum value at $x=1$ . Now evaluate the maximum value at $x=1$ from (1).

  $\begin{array}{c}{\left[f\left(1\right)\right]}_{\max }=2\left(1\right)-{\left(1\right)}^2\\ =1\end{array}$