Maths · Limits, continuity and differentiability
=\frac{\mathbf{1}}{\mathbf{2}}[\boldsymbol{f}(\boldsymbol{x})+\boldsymbol{f}(\bo
\( \operatorname{Let} \boldsymbol{f}\left(\frac{\boldsymbol{x}+\boldsymbol{y}}{\mathbf{2}}\right)=\frac{\mathbf{1}}{\mathbf{2}}[\boldsymbol{f}(\boldsymbol{x})+\boldsymbol{f}(\boldsymbol{y})] \) for real \( x \) and \( y . \) If \( f^{\prime}(0) \) exists and equals -1 and \( f(0)=1 \) then the value of \( f(2) \) is
- A. 1
- B. -
- C. \( 1 / 2 \)
- D. 2
Step-by-step solution
The functional equation is Jensen's equation, which implies f is affine: f(x) = ax + b. Using f(0)=1 gives b=1. f'(0)=a = -1, so f(x)= -x+1. Thus f(2)= -2+1 = -1.
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