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(A) Given the functional equation $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$.This is a Jensen's functional equation, which implies that $f(x)$

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-$1$

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(A) Given the functional equation $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$.This is a Jensen's functional equation, which implies that $f(x)$ is a linear function of the form $f(x) = ax + b$.We are given $f(0) = 1$.Substituting $x = 0$ into $f(x) = ax + b$, we get $f(0) = a(0) + b = 1$, so $b = 1$.Now, $f(x) = ax + 1$.Differentiating with respect to $x$, we get $f^{\prime}(x) = a$.We are given $f^{\prime}(0) = -1$.Since $f^{\prime}(x) = a$ is a constant, $f^{\prime}(0) = a = -1$.Thus, the function is $f(x) = -x + 1$.To find $f(2)$, substitute $x = 2$ into the function:$f(2) = -(2) + 1 = -1$.

Let $f: R \rightarrow R$ be defined by $f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}$ for all $x$ and $y$. If $f^{\prime}(0)$ exists and equals $-1$ and $f(0)=1$, then $f(2)=$

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