Questions: Sketch the interval (a, b) on the x-axis with the point c inside. Then find the largest value of δ>0 such that for all x, 0<x-c<δ implies a<x<b. a=1/8, b=1/3, c=1/4

Sketch the interval (a, b) on the x-axis with the point c inside. Then find the largest value of δ>0 such that for all x, 0<x-c<δ implies a<x<b.
a=1/8, b=1/3, c=1/4
Transcript text: Sketch the interval $(a, b)$ on the $x$-axis with the point c inside. Then find the largest value of $\delta>0$ such that for all $\mathrm{x}, 0<|\mathrm{x}-\mathrm{c}|<\delta$ implies $\mathrm{a}<\mathrm{x}<\mathrm{b}$. \[ a=\frac{1}{8}, b=\frac{1}{3}, c=\frac{1}{4} \]
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Solution

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Solution Steps

Step 1: Identify the interval and point

Given the interval \((a, b)\) on the \(x\)-axis with the point \(c\) inside: \[ a = \frac{1}{8}, \quad b = \frac{1}{3}, \quad c = \frac{1}{4} \]

Step 2: Determine the distance from \(c\) to the endpoints of the interval

Calculate the distance from \(c\) to \(a\) and \(b\): \[ \delta_1 = c - a = \frac{1}{4} - \frac{1}{8} = \frac{1}{8} \] \[ \delta_2 = b - c = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \]

Step 3: Find the largest value of \(\delta\)

The largest value of \(\delta\) is the minimum of \(\delta_1\) and \(\delta_2\): \[ \delta = \min\left(\frac{1}{8}, \frac{1}{12}\right) = \frac{1}{12} \]

Final Answer

The largest value of \(\delta > 0\) such that for all \(x\), \(0 < |x - c| < \delta\) implies \(a < x < b\) is: \[ \delta = \frac{1}{12} \]

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