Question

From the given table, the value of f(153) using Lagrange's Interpolation Formula is
150
152
154
154
12.247
12329
12.410
12.490​

Answer 1

Answer:

12.410 i hope it's help you

Answer 2

Correct Question: Given table is:

           x :   $x_0=$ 150        $x_1=$ 152           $x_2=$ 154        $x_3=$ 156

f(x) = $\sqrt{x}$ :   $y_0=$12.247    $y_1=$ 12.329      $y_2=$ 12.410    $y_3=$ 12.490

Find the value of $f(153)$ using Lagrange's interpolation formula.

Answer:

The value of $f(153)$ is $0.8113$

Step-by-step explanation:

Lagrange's interpolation formula is given defined as

$y(x)=\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)} y_0 +\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)} y_1+\frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)} y_2$

          $+\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)} y_3$  ...... (1)

Substitute the value $153$ for $x$ in equation (1) as follows:

$y(153)=\frac{(153-x_1)(153-x_2)(153-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)} y_0 +\frac{(153-x_0)(153-x_1)(153-x_2)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)} y_1$

            $+\frac{(153-x_0)(153-x_1)(153-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)} y_2+\frac{(153-x_0)(153-x_1)(153-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)} y_3$  ...... (2)

Substitute the values of $x_0,x_1,x_2$, $x_3$ and $y_0,y_1,y_2,y_3$ in the equation (2) as follows:

$y(153)=\frac{(153-152)(153-154)(153-156)}{(150-152)(150-154)(150-156)} 12.247 +\frac{(153-150)(153-152)(153-154)}{(152-150)(152-154)(152-156)} 12.329$

            $+\frac{(153-150)(153-152)(153-156)}{(154-150)(154-152)(154-156)} 12.410+\frac{(153-150)(153-152)(153-154)}{(156-150)(156-152)(156-154)} 12.490$

$y(153)=\frac{(1)(-1)(-3)}{(-2)(-4)(-6)} 12.247 +\frac{(3)(1)(-1)}{(1)(-2)(-4)} 12.329$

             $+\frac{(3)(1)(-3)}{(4)(2)(-2)} 12.410+\frac{(3)(1)(-1)}{(6)(4)(2)} 12.490$

$y(153)=-0.7654 -4.6233+6.9806-0.7806$

$y(153)=0.8113$

Therefore, the value of $f(153)$ is $0.8113$.