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

in Calculus by Veteran (52.2k points)
edited by | 1.6k views

3 Answers

+13 votes
Best answer
by Veteran (432k points)
edited by
+23

just_bhavana 

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 ?? 

0

 Vicky rix

but how u derive option d from ur comment

+1
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.
+8 votes

Simplest Approach! 

by Junior (933 points)
0 votes
Let f(x) = 3x^2 and X0 = 2
Now, substitute for all the options. Only option (d) will come out to be true.
by (353 points)
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