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Chapter 1: Problem 43

Sketch the graph of the function and describe the interval(s) on which the function is continuous. $$ f(x)=\left\\{\begin{array}{ll}{x^{2}+1,} & {x<0} \\ {x-1,} & {x \geq 0}\end{array}\right. $$

Short Answer

Expert verified
Following the piecewise graphing rules, the graph can be plotted for the given intervals. The function is continuous at x=0 because the limit from both sides (positive and negative) matches the function value at x=0. It is also continuous for all values of x except possibly at x=0, but after evaluation, the function is found to be continuous over all real numbers.

Step by step solution

01

Sketch the first function

The function is broken into two intervals, so we need to sketch each separately. For x < 0, the function is given as \(f(x) = x^{2} + 1\). This is simply a quadratic function with the parabola opening upwards and shifted one unit upwards. But since the function is defined for x < 0, erase the right half of the quadratic parabola, that is, for x < 0.
02

Sketch the second function

For \(x \geq 0\), the function is given by \(f(x) = x - 1\). This is a linear function with slope 1 and y-intercept at -1. Following the definition, draw this function for \(x \geq 0\).
03

Verify continuity of the function

The function will be continuous if the end of the first function \(f(x) = x^{2} + 1\) for x < 0, joins the start of the second function \(f(x) = x - 1\) for \(x \geq 0\). Plugging 0 into both functions gives \(f(0) = 1\) in both cases, so the function is continuous over all given intervals.

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