
If y = f(x) is an odd differentiable function defined on $\left( { - \infty ,\infty } \right)$ such that ${f^\prime }(3) = - 2$ then ${f^\prime }( - 3)$is ?
Answer
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Hint: The given function is odd differentiable . For an odd differentiable function ${f^\prime }( - x) = - {f^\prime }(x)$and we are given${f^\prime }(3) = - 2$. Using the above property we can find ${f^\prime }( - 3)$
Complete step-by-step answer:
We are given that y = f(x) is an odd differentiable function.
And when a function is odd differentiable then we know that .${f^\prime }( - x) = - {f^\prime }(x)$.
We are given that ${f^\prime }(3) = - 2$
And we are asked for the value of
Using the above property,
$
\Rightarrow {f^\prime }( - 3) = - {f^\prime }(3) \\
\Rightarrow {f^\prime }( - 3) = - ( - 2) \\
\Rightarrow {f^\prime }( - 3) = 2 \\
$
So now we have the value of ${f^\prime }( - 3) = 2$.
Note: If the function f(x) is even differentiable then f( - x ) = f(x)
A function is differentiable at a point when there's a defined derivative at that point. This means that the slope of the tangent line of the points from the left is approaching the same value as the slope of the tangent of the points from the right.
Complete step-by-step answer:
We are given that y = f(x) is an odd differentiable function.
And when a function is odd differentiable then we know that .${f^\prime }( - x) = - {f^\prime }(x)$.
We are given that ${f^\prime }(3) = - 2$
And we are asked for the value of
Using the above property,
$
\Rightarrow {f^\prime }( - 3) = - {f^\prime }(3) \\
\Rightarrow {f^\prime }( - 3) = - ( - 2) \\
\Rightarrow {f^\prime }( - 3) = 2 \\
$
So now we have the value of ${f^\prime }( - 3) = 2$.
Note: If the function f(x) is even differentiable then f( - x ) = f(x)
A function is differentiable at a point when there's a defined derivative at that point. This means that the slope of the tangent line of the points from the left is approaching the same value as the slope of the tangent of the points from the right.
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