Question

Values of f(x) for values of x are given as

f(1) = 4, f(2) = 5, f(7) = 5, f(8) = 4

Find f(6) and also the value of x for which f(x) is maximum or minimum.

Answer 1

Here the intervals are unequal.

$$\begin{matrix} x_0=1, & x_1=2, & x_2=7, & x_3=8 \\ y_0=4, & y_1=5, & y_2=5, & y_3=4 \end{matrix}$$

By Lagrange’s interpolation formula we have

$$y=f(x)=\dfrac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}\times y_0$$

$$+\dfrac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}\times y_1$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}\times y_2$$

$$+\dfrac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}\times y_3$$

Put $x=6$

$$f(6)=\dfrac{(6-2)(6-7)(6-8)}{(1-2)(1-7)(1-8)}\times 4$$

$$+\dfrac{(6-1)(6-7)(6-8)}{(2-1)(2-7)(2-8)}\times 5$$

$$+\dfrac{(6-1)(6-2)(6-8)}{(7-1)(7-2)(7-8)}\times5$$

$$+\dfrac{(6-1)(6-2)(6-7)}{(8-1)(8-2)(8-7)}\times 4$$

$$=\dfrac{17}{3}$$

$f(6)=\dfrac{17}{3}$

Let $f(x)=ax^2+bc+c$

$$\begin{alignedat}{2} a+b+c= 4 \\ 4a+2b+c=5 \\ 49a+7b+c=5 \\ 64a+8b+c=4 \end{alignedat}$$

$$\begin{alignedat}{2} a+b+c= 4 \\ 3a+b=1 \\ 15a+b=-1 \\ 64a+8b+c=4 \end{alignedat}$$

$a=-\dfrac{1}{6}, b=\dfrac{3}{2}, c=\dfrac{8}{3}$

$$f(x)=-\dfrac{1}{6}x^2+\dfrac{3}{2}x+\dfrac{8}{3}$$

$x_v=-\dfrac{\dfrac{3}{2}}{2(-\dfrac{1}{6})}=\dfrac{9}{2}$

$y_v=f(\dfrac{9}{2})=-\dfrac{1}{6}(\dfrac{9}{2})^2+\dfrac{3}{2}(\dfrac{9}{2})+\dfrac{8}{3}=\dfrac{145}{24}$

f(x) is maximum at $x=\dfrac{9}{2}$