The identity attribute f(x) = x is constant in its domain. If f(x) and g(x) are both continuous at x = c, so is f(x) + g(x) at x = c. If f(x) and g(x) are both constant at x = c, so is f(x) * g(x) at x = c. If f(x) and g(x) are both continuous at x = c, and g(x) # 0, then f(x) / g(x) is consistent at x = c. If f(x) is continuous at x = c, and also g(x) is constant at x = f(c), then the complace g(f(x)) is constant at x = c.


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Proof: Suppose f(x) = x. Then, provided any > 0 choose = / 2. Then, if | x - c | it means that | f(x) - f(c) | = | x - c | . Hence, the identification function is indeed continuous. Was it really important to take = / 2 ? The sum of constant attributes is constant complies with directly from the triangle inequality. Take any > 0. Tright here exists 1 > 0 such that whenever before | x - c | 1 we recognize that | f(x) - f(c) | (because f is continuous at c). Tbelow likewise exists 2 > 0 such that whenever before | x - c | 2 we know that | g(x) - g(c) | (bereason g is consistent at c). But then, if we let = min(1 , 2), we have: if | x - c | then | (f(x) + g(x)) - (f(c) + g(c)) | ≤ | f(x) - f(c) | + | g(x) - g(c) | + = 2 That finishes the proof. (That we do not gain a simple must not bvarious other us any kind of more). The product of 2 consistent attributes is aget consistent, which complies with from a straightforward trick. We will only look at the trick associated, and leave the details to the reader: | f(x) g(x) - f(c) g(c) | = | f(x) g(x) - f(x) g(c) + f(x) g(c) - f(c) g(c) | ≤ | f(x) | | g(x) - g(c) | + | g(c) | | f(x) - f(c) | With this trick the rest of the proof have to not be also hard. A comparable trick functions for the quotient. Here is the idea: | f(x) / g(x) - f(c) / g(c) | = | 1 / g(x) g(c) | | f(x) g(c) - f(c) g(x) | Can you watch just how to proceed ? Adding and subtracting will certainly assist aacquire. As for composition of functions, we have to proceed somewhat different: We recognize that f(x) is constant at c, and also g(x) is consistent at f(c). Because of this, provided any > 0 tbelow exists 1 > 0 such that whenever before | t - d | 1 then | g(t) - g(d) | Tright here also exists 2 > 0 such that if | x - c | 2 then | f(x) - f(c) | 1 . (Keep in mind that we have actually reput the usual by 1 here) Now let = min(1 , 2) and also substitute t = f(x) and also d = f(c). We have: if | x - c | then | f(x) - f(c) | 1 and then | g(f(x)) - g(f(c)) | In other words, f(g(x)) is constant at x = c.


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Interenergetic Real Analysis, ver. 2.0.1(c) (c) 1994-2018, Bert G. Wachsmuth Page last modified: Mar 2, 2018