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The formula used to compute an approximation for the second derivative of a function $f$ at a point $X_0$ is

1. $\dfrac{f(x_0 +h) + f(x_0 – h)}{2}$

2. $\dfrac{f(x_0 +h) - f(x_0 – h)}{2h}$

3. $\dfrac{f(x_0 +h) + 2f(x_0) + f(x_0 – h)}{h^2}$

4. $\dfrac{f(x_0 +h) - 2f(x_0) + f(x_0 – h)}{h^2}$

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edited

The below formula is also correct ...right ???

Because Left-Hand-Derivative = Right Hand Derivative //Else f ' (x) doesnot exist at that point.

Similiarly Left hand second derivative should be equal to right hand second derivative ..right ??? In that case below formula will also be correct ..right ??

but how u derive option d from ur comment

Hello vicky rix

Yes that's also true. in your final answer , we can say that $f''(x)$=$f''(x+h)$=$f''(x-h)$

so when you substitute $x$ by $x-h$ you will get the original answer.

Simplest Approach!

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Let f(x) = 3x^2 and X0 = 2
Now, substitute for all the options. Only option (d) will come out to be true.