Let f: R rightarrow R be a bijection. A curve | vedclass

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(A) Given $y=f(x)$. The point $P$ is $(\alpha, 1)$.Equation of the tangent at $P(\alpha, 1)$ is $y-1=f^{\prime}(\alpha)(x-\alpha)$.Setting $y=0$, the

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$x-y=0$

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(A) Given $y=f(x)$. The point $P$ is $(\alpha, 1)$.Equation of the tangent at $P(\alpha, 1)$ is $y-1=f^{\prime}(\alpha)(x-\alpha)$.Setting $y=0$, the $X$-intercept $A$ is $\alpha - \frac{1}{f^{\prime}(\alpha)}$. So $A = (\alpha - \frac{1}{f^{\prime}(\alpha)}, 0)$.Equation of the normal at $P(\alpha, 1)$ is $y-1=-\frac{1}{f^{\prime}(\alpha)}(x-\alpha)$.Setting $y=0$, the $X$-intercept $B$ is $\alpha + f^{\prime}(\alpha)$. So $B = (\alpha + f^{\prime}(\alpha), 0)$.The point $C$ is $(\alpha, 0)$.Then $AC = |\alpha - (\alpha - \frac{1}{f^{\prime}(\alpha)})| = \frac{1}{f^{\prime}(\alpha)}$ and $CB = |(\alpha + f^{\prime}(\alpha)) - \alpha| = f^{\prime}(\alpha)$.We want to minimize $AC+CB = \frac{1}{f^{\prime}(\alpha)} + f^{\prime}(\alpha)$.By $AM$-$GM$ inequality, $\frac{1}{f^{\prime}(\alpha)} + f^{\prime}(\alpha) \geq 2 \sqrt{\frac{1}{f^{\prime}(\alpha)} \cdot f^{\prime}(\alpha)} = 2$.The minimum occurs when $f^{\prime}(\alpha) = 1$.The slope of the tangent at $P$ is $f^{\prime}(\alpha) = 1$.$A$ line parallel to the tangent must have slope $1$. The line $x-y=0$ has slope $1$.

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