Question

if f(0)=12,f(3)=6 and f(4)=8,then linear interpolation function f(x)=?
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Answer 1

Linear interpolation function f(x) = x²-5x+12.

Given,

f(0)=12

f(3)=6

f(4)=8

To find,

Linear interpolation function f(x)

Solution:

By the given conditions,

• ω(x) = ( x – 0 ) ( x – 3 ) ( x – 4 )
• ω(x) = x(x – 3)(x – 4 )

It gives,

• ω'(x) = (x – 3)(x – 4 ) + x(x – 4 ) + x(x – 3)
• ω'(0) = 12
• ω'(3) = – 3
• ω'(4) = 4

Hence, the required polynomial,

• f(x) = ω(x) Σ f(xr) / (x-xr) ω'(xr)
• f(x) = x(x-3) (x-4) { [ 12/ 12(x-0) ] + [ 6/-3(x-3) ] + [8/4(x-4) ]}
• f(x) = x(x-3) (x-4) { [ 1/x ] - [ 2/(x-3) ] + [2/(x-4) ]}
• f(x) = x(x-3) (x-4) - 2x (x-4) + 2x(x-3)
• f(x) = x²-5x+12

Linear interpolation function f(x) = x²-5x+12.

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Answer 2

Answer:

linear interpolation function f(x)= x²-5x+12

Step-by-step explanation:

Given :

f(0)=12,

f(3)=6

and f(4)=8

To Find:

linear interpolation function f(x)=?

Solution :-

here  it is given that

f(0)=12,

f(3)=6

and f(4)=8

Now , ω(x)

= (x-0)(x-3)(x-4)

= x(x-3)(x-4)

∴ω'(x)

=(x-3)(x-4)+x(x-4)+x(x-3)

ω'(0) = 12

ω'(3) = -3

ω'(4) = 4

Now the required polynomial

= ω(x)∑ $\frac{f(x_{r}) }{(x-x_{r} ) w' (x_{r}) }$

= x(x-3)(x-4)[ $[\frac{12}{12 (x-0)} +\frac{6}{-3(x-3)} +\frac{8}{4(x-4)}]$

= x(x-3)(x-4)$[\frac{1}{x} -\frac{2}{(x-3)} +\frac{2}{(x-4)}]$

=(x-3)(x-4)-2x(x-4)+2x(x-3)

= x²-5x+12

linear interpolation function f(x)= x²-5x+12

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