The shortest distance of the point $\left(\frac{1}{2}, 1\right)$ from the curve $y=|x-1|+|x+1|$ is

We need to find the shortest distance from the point $\left(\frac{1}{2}, 1\right)$ to the curve $y = |x-1| + |x+1|$.

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To sketch $y = |x-1| + |x+1|$, we need to analyze where each absolute value expression changes sign:

$|x-1|$ changes at $x = 1$

$|x+1|$ changes at $x = -1$

This gives us three intervals to consider:

For $x < -1$:

Both $(x-1)$ and $(x+1)$ are negative

So $|x-1| = -(x-1) = -x+1$ and $|x+1| = -(x+1) = -x-1$

Therefore: $y = (-x+1) + (-x-1) = -2x$

For $-1 \leq x < 1$:

$(x-1)$ is negative, $(x+1)$ is non-negative

So $|x-1| = -(x-1) = -x+1$ and $|x+1| = x+1$

Therefore: $y = (-x+1) + (x+1) = 2$

For $x \geq 1$:

Both $(x-1)$ and $(x+1)$ are non-negative

So $|x-1| = x-1$ and $|x+1| = x+1$

Therefore: $y = (x-1) + (x+1) = 2x$

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$y = \begin{cases} -2x & \text{if } x < -1 \ 2 & \text{if } -1 \leq x < 1 \ 2x & \text{if } x \geq 1 \end{cases}$

This creates a V-shaped curve with a flat bottom segment.

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The point $\left(\frac{1}{2}, 1\right)$ has x-coordinate $\frac{1}{2}$, which falls in the interval $[-1, 1)$.

In this interval, the curve is simply $y = 2$ (a horizontal line).

Since our point is at $\left(\frac{1}{2}, 1\right)$ and the curve directly above it is at $\left(\frac{1}{2}, 2\right)$, the shortest distance is the vertical distance between these points.

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Distance = $|2 - 1| = 1$

The horizontal segment $y = 2$ for $-1 \leq x < 1$ is the closest part of the curve to our point. Any other point on the slanted portions of the curve would be farther away. The perpendicular distance from a point to a horizontal line is simply the difference in y-coordinates.

Therefore, the shortest distance is 1.