Suppose F(x) = X2 + 1 X ≤ 2 3x 1 Given That

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Suppose f(x) = x2 + 1 x ≤ 2 3x 1 Given that 2 < x < 4 9 2x 4 ≤ x On each of the intervals (-∞, 2), (2, 4), and (4,∞), f is a polynomial function, and so, according to the reading in section 2.4, f is continuous on each of these intervals. Is f either left or right continuous (or both) at x = 2 Is f either left or right continuous (or both) at x = 4 Draw the graph of f

 

Answer No.1

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F(x) = x2 + 1 x ≤ 2 3x 1 2 < x < 4 9 2x 4 ≤ x at x = 2 lt (x--> 2-) f(x) = x2 + 1 = 2^2 + 1 = 5 f(2) = x2 + 1 = 5 lt (X--> 2+) f(x) = 3x 1 = 3*2 - 1 = 5 => x is continous at 2 lt (x--> 2-) f(x) = f(2) ....=> f is left continuous at 2 lt (X--> 2+) f(x) = f(2).......=> f is right continuous at 2 at x = 4 lt (x--> 4-) f(x) = 3x 1 = 3*4 - 1 = 11 f(4) = 9 2x = 1 lt (X--> 4+) f(x) = 9 2x = 1 => x is not continous at 4 t (x--> 4-) f(x) not equal to f(4) ....=> f is not left continuous at 4 lt (X--> 4+) f(x) = f(4).......=> f is right continuous at 4

Answer No.2

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At x=2 left hand lim f(x) = 2^2 +1 =5 right hand lim f(x) = 3*2 -1 =5 thus it is both left and right continuous at x=2 at x=4 left hand lim f(x) = 3*4 -1 =11 right hand lim f(x) =9-2*4 =1 thus it is left continuous at x=4

Answer No.3

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AT x=2: left limit of 'f' is 2^2+1=5. right limit of 'f' is 3*2-1=5 f(2)=5 hence 'f' is both left and right continuous at =2. at x=4: left limit is 3*4-1=11 right limit is 9-2*4=1. f(4)=1 'f' is right continuous but not left continuous.

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