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Question

Answers

\[\begin{align}

& A.12.2 \\

& B.11.2 \\

& C.10.2 \\

& D.15.2 \\

\end{align}\]

Answer
Verified

Before solving the question, we should know how the newton’s forward formula is used. If\[f(x)=f(a)+m\left( f(b)-f(a) \right)\] where a and b are two integers, \[a\le x\le b\], \[m=\dfrac{x-a}{h}\] where \[h=b-a\].

From the question, it was given that \[f(1)=10,f(2)=14\]. Now we should find the value of \[f(1.3)\] using newton’s forward formula.

Now let us compare \[f(1)=10,f(2)=14\] with \[f(a),f(b)\]. Then the value of a is equal to 1 and the value of b is equal to 2.

We know that \[h=b-a\].

Let us consider

\[h=b-a....(3)\]

Now let us substitute equation (1) and equation (2) in equation (3), then we get

\[\begin{align}

& \Rightarrow h=2-1 \\

& \Rightarrow h=1....(4) \\

\end{align}\]

From the question, it is clear that we should find the value of \[f(1.3)\].

Now we will let us assume \[f(x)\] with \[f(1.3)\].

Then, we get

\[\Rightarrow x=1.3....(5)\]

We know that \[m=\dfrac{x-a}{h}\].

Let us consider

\[m=\dfrac{x-a}{h}......(6)\]

So, let us substitute equation (5), equation (1) and equation (4) in equation (6), then we get

\[\begin{align}

& \Rightarrow m=\dfrac{1.3-1}{1} \\

& \Rightarrow m=0.3......(7) \\

\end{align}\]

We know that \[f(x)=f(a)+m\left( f(b)-f(a) \right)\] where a and b are two integers, \[a\le x\le b\], \[m=\dfrac{x-a}{h}\] where \[h=b-a\].

Let us consider

\[f(x)=f(a)+m\left( f(b)-f(a) \right).....(8)\]

We know that \[f(1)=10,f(2)=14\].

So, let us consider

\[\begin{align}

& f(1)=10.....(9) \\

& f(2)=14.....(10) \\

\end{align}\]

Now we will substitute equation (3), equation (4), equation (5), equation (6), equation (7), equation (9) and equation (10) in equation (8), then we get

\[\begin{align}

& \Rightarrow f(1.3)=10+\left( 0.3 \right)\left( 14-10 \right) \\

& \Rightarrow f(1.3)=10+\left( 0.3 \right)\left( 4 \right) \\

& \Rightarrow f(1.3)=10+1.2 \\

& \Rightarrow f(1.3)=11.2......(11) \\

\end{align}\]

Now from equation (11), it is clear that the value of \[f(1.3)\] is equal to 11.2.