course Mth 271 ^䚗\UզКassignment #011
......!!!!!!!!...................................
11:33:14 1.6.16 (was 1.6.14 intervals of cont for (x-3)/(x^2-9) What are the intervals of continuity for the given function?
......!!!!!!!!...................................
RESPONSE --> continuous on the intervals of (-~,0) (0, ~) confidence assessment: 1
.................................................
......!!!!!!!!...................................
11:34:48 The function is undefined where x^2 - 9 = 0, since division by zero is undefined. x^2 - 9 = 0 when x^2 = 9, i.e., when x = +-3. So the function is continuous on the intervals (-infinity, -3), (-3, 3) and (3, infinity). The expression (x - 3) / (x^2 - 9) can be simplified. Factoring the denominator we get (x - 3) / [ (x - 3) ( x + 3) ] = 1 / (x + 3). This 'removes' the discontinuity at x = +3. However in the given fom (x-3) / (x^2 + 9) there is a discontinuity at x = -3. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 1
.................................................
......!!!!!!!!...................................
11:35:23 1.6.24 (was 1.6.22 intervals of cont for |x-2|+3, x<0; x+5, x>=0 What are the intervals of continuity for the given function?
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 1
.................................................
......!!!!!!!!...................................
11:36:19 The graph of y = |x-2|+3 is translated 2 units in the x direction and 3 in the y direction from the graph of y = |x|. It forms a V with vertex at (2, 3). The given function follows this graph up to x = 0. It has slope -1 and y-intercept at y = | 0 - 2 | + 3 = 5. The graph then follows y = x + 5 for all positive x. y = x + 5 has y-intercept at y = 5. From that point the graph increases along a straight line with slope 1. So the graph of the given function also forms a V with vertex at (0, 5). Both functions are continuous up to that point, and both continuously approach that point. So the function is everywhere continuous. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 1
.................................................