Question

If $y = f(x)$ is an odd differentiable function defined on $\left( { - \infty ,\infty } \right)$ such that $f'\left( 3 \right) = - 2$ then $f'\left( { - 3} \right)$ equals
A) 4
B) 2
C) -2
D) 0

Answer 1

It is given in the question that If $y = f(x)$ is an odd differentiable function defined on $\left( { - \infty ,\infty } \right)$ such that $f'\left( 3 \right) = - 2$ then $f'\left( { - 3} \right)$ is
First, it is given that the function $y = f(x)$ is an odd function.
$\therefore f\left( { - x} \right) = - f\left( x \right)$
Then after, we will differentiate the above equation and finally we will get the answer.

Complete step by step solution:
It is given in the question that If $y = f(x)$ is an odd differentiable function defined on $\left( { - \infty ,\infty } \right)$ such that $f'\left( 3 \right) = - 2$ then $f'\left( { - 3} \right)$ is
It is given that function $y = f(x)$ is odd function.
$\therefore f\left( { - x} \right) = - f\left( x \right)$
Now, differentiate the above equation with respect to x, we get,
$\therefore \left( { - 1} \right)f'\left( { - x} \right) = - f'\left( x \right) \\\ \therefore f'\left( { - x} \right) = f'\left( x \right) \\\$
Now, let x = 3, we get,
$\therefore f'\left( { - 3} \right) = f'\left( 3 \right)$
Since, it is given in the question that $f'\left( 3 \right) = - 2$

$\therefore f'\left( { - 3} \right) = - 2$

Note:
Even Function: Let f be a real valued function of a real variable. Then f is even in the following equation holds for all x such that x and -x in the domain of f.
$f\left( x \right) = f\left( { - x} \right)$
$f\left( x \right) + f\left( x \right) = 0$
Odd Function: Let f be a real valued function of a real variable. Then f is odd in the following equation holds for all x such that x and -x in the domain of f.
$f\left( { - x} \right) = - f\left( x \right)$
$f\left( { - x} \right) + f\left( x \right) = 0$