A calculus problem by Rohan Rao
Calculus
Level
pending
The function f(x) = 3x+1 is continuous at which value of x?
1
4
0
2
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1 solution
we have f(1)=4 and f(x)-f(1)= 3x+1-4 = 3(x-1) We will now attempt to make the numerical value of this difference smaller than any pre-assigned positive number, say 0.001 i) Let x>1 so that 3(x-1)is positive and is therefore the numerical value of f(x)-f(1). Now | f(x)-f(1)| = 3(x-1_ < 0.001 if x-1<0.001/3, i.e., if x<1+0.001/3 similarly, let it be negative then the above value becomes |f(x)-f(1)|=3(x-1)<0.001 if 1-x<0.001/3 i.e. if 1-0.001/3<x combining the above two results we get | f(x)-f(1)| <0.001 for all those value of x for which 1-0.001/3<x<1+0.001/3 The test of continuity for 1 is thus satisfied for the particular value of 0.001 of epsilon e. Hence f is continuous at 1