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

  • 🚩 Edit necessary | 👮 Arjun
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3 Answers

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For a function $f(x)$ let $f'(x)$ denote its derivative. 

By the definition of derivative, $\displaystyle f'(x) = \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}.$

By substituting $f$ with $f'$ in the above equation (the same formula should hold for any function and so we can do this), we get

$\displaystyle f''(x) =   \lim_{h \to 0} \dfrac{f'(x+h) - f'(x)}{h}$

$\implies \displaystyle f''(x) =   \lim_{h \to 0} \dfrac{\displaystyle  \lim_{h \to 0} \dfrac{f(x+h + h) - f(x+h)}{h}   - \lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}}{h}$

$\implies \displaystyle f''(x) =   \lim_{h \to 0} \dfrac{f(x+2h) - f(x+h)   -{f(x+h) + f(x)}}{h^2}$

$\implies \displaystyle f''(x) =   \lim_{h \to 0} \dfrac{f(x+2h) - 2f(x+h)  + f(x)}{h^2}$

At $x = x_0-h$

$\displaystyle f''(x) =   \lim_{h \to 0} \dfrac{f(x_0+h) - 2f(x_0)  + f(x_0-h)}{h^2}$

Option $D.$

Ref: http://en.wikipedia.org/wiki/Second_derivative

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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.
Answer:

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