Mathematics

Mathematics

Problem

The function hh is defined by h(x)={1x−4,x<45−x,x≥4.h(x)=\begin{cases}\dfrac{1}{x-4}, & x<4\\[6pt] \sqrt{5-x}, & x\ge4.\end{cases} On which of the following intervals is hh continuous? a) (−∞,4)∪[4,5](-\infty,4)\cup[4,5] b) (−∞,4)∪(4,5](-\infty,4)\cup(4,5] c) (−∞,4)∪(4,5)(-\infty,4)\cup(4,5) d) (−∞,4)∪(4,∞)(-\infty,4)\cup(4,\infty)
Intuitive explanation

The key to this problem is understanding that a piecewise function is continuous everywhere except possibly at the boundary where the definition changes. You need to check two things: whether each piece is continuous on its own domain, and whether the two pieces match up at x=4x=4 (do the left and right limits agree, and do they equal the function value?). The piece on (−∞,4)(-\infty, 4) has a vertical asymptote that you should identify, and the piece on [4,∞)[4, \infty) is a square root with its own domain restriction—so hh will be continuous only on intervals that avoid these problem points.

Intuitive explanation

The key to this problem is understanding that a piecewise function is continuous everywhere except possibly at the boundary where the definition changes. You need to check two things: whether each piece is continuous on its own domain, and whether the two pieces match up at x=4x=4 (do the left and right limits agree, and do they equal the function value?). The piece on (−∞,4)(-\infty, 4) has a vertical asymptote that you should identify, and the piece on [4,∞)[4, \infty) is a square root with its own domain restriction—so hh will be continuous only on intervals that avoid these problem points.

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