Let f(x) = x^m, where m is a non-negative int | vedclass

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(C) Given $f(x) = x^m$, where $m \geq 0$ and $m \in \mathbb{Z}$.First, find the derivative $f^{\prime}(x) = m x^{m-1}$.The given condition is $f^{\pri

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$2$

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(C) Given $f(x) = x^m$, where $m \geq 0$ and $m \in \mathbb{Z}$.First, find the derivative $f^{\prime}(x) = m x^{m-1}$.The given condition is $f^{\prime}(a+b) = f^{\prime}(a) + f^{\prime}(b)$.Substituting the derivative, we get $m(a+b)^{m-1} = m a^{m-1} + m b^{m-1}$.If $m=0$, $f(x) = 1$, so $f^{\prime}(x) = 0$. The equation becomes $0 = 0+0$, which is true, but usually, we consider non-trivial cases for $m$. Let's check other values.If $m=1$, $f^{\prime}(x) = 1$. The equation becomes $1 = 1+1$, which is $1=2$ (False).If $m=2$, $f^{\prime}(x) = 2x$. The equation becomes $2(a+b) = 2a + 2b$, which simplifies to $2a+2b = 2a+2b$ (True).If $m=3$, $f^{\prime}(x) = 3x^2$. The equation becomes $3(a+b)^2 = 3a^2 + 3b^2$, which simplifies to $a^2 + 2ab + b^2 = a^2 + b^2$, implying $2ab = 0$. Since $a, b > 0$, this is False.Thus, the value of $m$ is $2$.

Let $f(x) = x^m$, where $m$ is a non-negative integer. The value of $m$ such that the equality $f^{\prime}(a+b) = f^{\prime}(a) + f^{\prime}(b)$ is valid for all $a, b > 0$ is

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