Chapter EP, Problem 5.1.12EP

To determine

To decide whether the function has a maximum or minimum value and to find the maximum or minimum value of the given function. Also to state the domain and range for the given function.

Answer to Problem 5.1.12EP

The given function has a minimum value.

The minimum value is $-\frac{1}{4}$.

The domain is all real numbers, that is, $\left\{-\infty <x<\infty \right\}$ and range is $\left\{f\left(x\right)|f\left(x\right)\ge -\frac{1}{4}\right\}$.

Explanation of Solution

Given information:

The function provided is $f\left(x\right)=x^2-5x+6$.

Formula used:

For the quadratic function $f\left(x\right)=a{x}^2+bx+c$

Formula to compute the x -coordinate of the vertex is $x=-\frac{b}{2\times a}$ .

‘a’ and ‘b’ are coefficients $x^2$ and $x$ respectively. The corresponding value of ‘y’ for the vertex would be the maximum or minimum.

Explanation :

For the function $f\left(x\right)=x^2-5x+6$ , the coefficients are $a=1$ , $b=-5$ and $c=6$.

Since the coefficient of ‘a’ is positive, the curve of this function is upward facingand hence the function has a minimum value.

x -coordinate of the vertex is

  $x=-\frac{b}{2\times a}$

Here, $a=1$ , $b=-5$ and $c=6$.

Formula for axis of symmetry.

  $\begin{array}{c}x=\frac{5}{2\times \left(1\right)}\\ =\frac{5}{2}\end{array}$

Vertex can be found out by putting the value of x computed in the axis of symmetry in the original function. This will give a value of y that will be the minimum.

  $\begin{array}{c}y={\left(\frac{5}{2}\right)}^2-5\left(\frac{5}{2}\right)+6\\ =\frac{25}{4}-\frac{25}{2}+6\\ =-\frac{1}{4}\end{array}$

Thus, the minimum value is $-\frac{1}{4}$.

Since x can take up any real values, the domainis $\left\{-\infty <x<\infty \right\}$.

Since the range is the values of y, it can only take up values lesser than or equal to the minimum. That is, $\left\{f\left(x\right)|f\left(x\right)\ge -\frac{1}{4}\right\}$.