Q. 75

Question

Curve Fitting Find the function $f (x) = a x^{3} + b x^{2} + c x + d$ for which $f (- 3) = - 112, f (- 1) = - 2, f (1) = 4$ , and $f (2) = 13$ .

Step-by-Step Solution

Verified

Answer

$f (x) = 3 x^{3} - 4 x^{2} + 5$

1Step 1. Given information

We have been given:

$f (- 3) = - 112, f (- 1) = - 2, f (1) = 4$ and $f (2) = 13$

We have to find $f (x) = a x^{3} + b x^{2} + c x + d$ .

2Step 2. Put x = - 3 in the function f ( x )   =   a x 3   +   b x 2   +   c x   +   d .

$a \cdot (- 3)^{3} + b \cdot (- 3)^{2} + c \cdot (- 3) + d = - 112 - 27 a + 9 b - 3 c + d = - 112$

3Step 3. Put x = - 1 in the function f ( x )   =   a x 3   +   b x 2   +   c x   +   d .

$a \cdot (- 1)^{3} + b \cdot (- 1)^{2} + c \cdot (- 1) + d = - 112 - a + b - c + d = - 2$

4Step 4. Put x = 1 in the function f ( x )   =   a x 3   +   b x 2   +   c x   +   d .

$a \cdot (1)^{3} + b \cdot (1)^{2} + c \cdot (1) + d = - 112 a + b + c + d = 4$

5Step 5. Put x = 2 in the function f ( x )   =   a x 3   +   b x 2   +   c x   +   d .

$a \cdot (2)^{3} + b \cdot (2)^{2} + c \cdot (2) + d = 13 8 a + 4 b + 2 c + d = 13$

6Step 6. Write the system of equations and corresponding augmented matrix

$\{\begin{cases} - 27 a + 9 b - 3 c + d = - 112 \\ - a + b - c + d = - 2 \\ a + b + c + d = 4 \\ 8 a + 4 b + 2 c + d = 13 \end{cases}$

The corresponding augmented matrix is:

$[\begin{array}{ccccc} - 27 & 9 & - 3 & 1 & - 112 \\ - 1 & 1 & - 1 & 1 & - 2 \\ 1 & 1 & 1 & 1 & 4 \\ 8 & 4 & 2 & 1 & 13 \end{array}]$

7Step 7. Perform elementary row operations

$R_{1} \leftrightarrow R_{3} [\begin{array}{ccccc} 1 & 1 & 1 & 1 & 4 \\ - 1 & 1 & - 1 & 1 & - 2 \\ - 27 & 9 & - 3 & 1 & - 112 \\ 8 & 4 & 2 & 1 & 13 \end{array}] R_{2} \to R_{2} + R_{1} R_{3} \to R_{3} + 27 R_{1} R_{4} \to R_{4} - 8 R_{1} [\begin{array}{ccccc} 1 & 1 & 1 & 1 & 4 \\ 0 & 2 & 0 & 2 & 2 \\ 0 & 36 & 24 & 28 & - 4 \\ 0 & - 4 & - 6 & - 7 & - 19 \end{array}] R_{3} \to R_{3} - 18 R_{2} R_{4} \to R_{4} + 2 R_{2} [\begin{array}{ccccc} 1 & 1 & 1 & 1 & 4 \\ 0 & 2 & 0 & 2 & 2 \\ 0 & 0 & 24 & - 8 & - 40 \\ 0 & 0 & - 6 & - 3 & - 15 \end{array}]$

8Step 8. Perform more row operations

$R_{2} \to \frac{1}{2} R_{2} R_{4} \to R_{4} + \frac{6}{24} R_{3} [\begin{array}{ccccc} 1 & 1 & 1 & 1 & 4 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 24 & - 8 & - 40 \\ 0 & 0 & 0 & - 5 & - 25 \end{array}] R_{3} \to \frac{1}{8} R_{3} R_{4} \to - \frac{1}{5} R_{4} : [\begin{array}{ccccc} 1 & 1 & 1 & 1 & 4 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 3 & - 1 & - 5 \\ 0 & 0 & 0 & 1 & 5 \end{array}]$

9Step 9. Solve for a,b,c and d

$a + b + c + d = 4 (1) b + d = 1 (2) 3 c - d = - 5 (3) d = 5 (4) d = 5 F r o m (3) : 3 c - 5 = - 5 c = 0 F r o m (2) : b + 5 = 1 b = - 4 F r o m (1) : a - 4 + 5 = 4 a = 3$

$Therefore, the function will be f (x) = 3 x^{3} - 4 x^{2} + 5 .$

Q. 74

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