GATE CSE 1996 | Question: 1.6

Question

The formula used to compute an approximation for the second derivative of a function $f$ at a point $x_0$ is

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

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

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

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

Answer 1

Simplest Approach! 

Answer 2

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

Comments

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.

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I got most of it, but while applying second derivative, why is it applied only on functions in the numerator but not on "h" in the denominator? Please help.

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@Deepa R Please see the answer now. In that step, we are not applying derivative, but rather applying the limit form of derivation to $f’(x)$

Answer 3

Let f(x) = 3x^2 and X0 = 2

Now, substitute for all the options. Only option (d) will come out to be true.