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Question : - Let f(x) be an even function such that f'(0) exists, find the value of f'(0).

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Answer 1

Given:-

f is an even function and f'(0) exists.

To Find:-

The value of f'(0)

Concept:-

• A function which satisfies f(x)= f(-x) is called even function.

• An even function is symmetric about y -axis.

• A function which satisfies f(x)= - f(-x) is called odd function. An odd function is symmetric about origin.

• If f is an even function, f' will be odd function and vice versa.

For an odd function f, f(0) = 0.

$\sf f'(0) = \begin{gathered}\displaystyle \sf \lim _{x \: to \: {0}^{ + }} \dfrac{f(h) - f(0)}{h} = \sf \displaystyle \sf \lim _{x \: to \: {0}^{ - }} \dfrac{f( - h) - f(0)}{ - h} \\ \\ \sf f \: is \: an \: even\:function \: f(h) = f( - h)\end{gathered}$

$\sf \displaystyle \sf \lim _{x \: to \: {0}^{ + }} \frac{f(h) - f(0)}{h} =</p><p> \sf \displaystyle \sf \lim _{x \: to \: {0}^{ - }} \frac{f( h) - f(0)}{ - h}x \: to \: 0+limh$

$\sf \begin{gathered} \\ \\ \sf ⇒ 2 \displaystyle \sf \lim _{x \: to \: {0}^{ + }} \dfrac{f(h) - f(0)}{h} = 0 \\ \\ \sf ⇒ 2f'(0) = 0 \\ \\ \sf ➾ f'(0) = 0\end{gathered}$