If \( f: R \rightarrow R \) defined as of \( f(x)=x^{2}+1 \) then minimum value of \( f(x) \) is
The given function is \[ f(x) = x^2 + 1 \] This is a quadratic function of the form \( ax^2 + bx + c \), where \( a = 1 > 0 \), indicating the parabola opens upwards. The minimum value of the function occurs at the vertex of the parabola. The formula for the \( x \)-coordinate of the vertex is: \[ x = -\frac{b}{2a} \] Here, \( b = 0 \) and \( a = 1 \), so: \[ x = -\frac{0}{2(1)} = 0 \] Substitute \( x = 0 \) into the function to find the minimum value: \[ f(0) = 0^2 + 1 = 1 \] Thus, the minimum value of \( f(x) \) is: \[ \boxed{1} \]
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