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Question: Discuss the continuity of the function where \(f\), where \(f\) is defined by \(f(x)=\left\\{ \be...

Discuss the continuity of the function where ff, where ff is defined by
f(x)=\left\\{ \begin{aligned} & 3,\text{ }if\text{ }0\le x\le 1 \\\ & 4,\text{ }if\text{ 1}x<3 \\\ & 5,\text{ }if\text{ 3}\le x\le 10\text{ } \\\ \end{aligned} \right.\text{ }

Explanation

Solution

Hint: Here we have to apply the condition for continuity that, if Left Hand Limit is equal to the Right Hand Limit which is the same as the function. i.e. limxa+f(x)=limxaf(x)=f(a)\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)=f(a), then the function is continuous. Also check at particular points where the function splits.

Complete step-by-step solution-
Here, consider the function ff defined by:
f(x)=\left\\{ \begin{aligned} & 3,\text{ }if\text{ }0\le x\le 1 \\\ & 4,\text{ }if\text{ 1}x<3 \\\ & 5,\text{ }if\text{ 3}\le x\le 10\text{ } \\\ \end{aligned} \right.\text{ }
So, here we have to check the continuity of the function f(x)f(x)
By definition we know that a function is said to be continuous in an open interval (a, b) if it is continuous at every point in the interval. For a closed interval, [a, b], f is continuous in (a, b), and
limxa+f(x)=f(a)limxbf(x)=f(b) \underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x)=f(a)\text{, }\underset{x\to {{b}^{-}}}{\mathop{\lim }}\,f(x)=f(b)\text{ }
Similarly, a function is said to be discontinuous at a point x=ax=a if
limxa+f(x)\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x) and limxaf(x)\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)exists but are not equal.
limxa+f(x)\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f(x) and limxaf(x)\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f(x)exists and are equal but not equal to f(a)f(a).
f(a)f(a) not defined.
We can solve this in different cases .First consider,
Case 1: Consider the interval 0x10\le x\le 1 . Here, the function value is f(x)=3f(x)=3, which is a constant function. Every constant function is continuous so we can say that
f(x)f(x) is continuous in the interval 0x10\le x\le 1.

Case 2: Let us consider the interval 1x<3\text{1}x<3. Here, the function value is f(x)=4f(x)=4, which is also a constant function. Therefore f(x)f(x) is continuous in the interval 1x<3\text{1}x<3.

Case 3: Here, consider the interval 3x10\text{3}\le x\le 10. In this interval the function value is f(x)=5f(x)=5 which is again a constant function. Hence, f(x)f(x) is continuous in the interval 3x10\text{3}\le x\le 10.

Case 4: Now, let us consider at x=1x=1.
We have the Left Hand Limit, limx1f(x)=3\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)=3 and the Right Hand Limit, limx1+f(x)=4\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)=4. So we can say that the Left Hand Limit is not equal to the right Hand Limit. i.e.
limx1f(x)limx1+f(x)\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,f(x)
Hence, f(x)f(x) is discontinuous at x=1x=1.

Case 5: Next, let us consider at x=3x=3.
Here, we have the Left Hand Limit, limx3f(x)=4\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f(x)=4 and the Right Hand Limit limx3+f(x)=5\underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f(x)=5. Therefore, the Left Hand Limit is not equal to the right Hand Limit. i.e.
limx3f(x)limx3+f(x)\underset{x\to {{3}^{-}}}{\mathop{\lim }}\,f(x)\ne \underset{x\to {{3}^{+}}}{\mathop{\lim }}\,f(x).
Hence, f(x)f(x) is discontinuous at x=3x=3.

The above cases describe the continuity of the function f(x)f(x).

Note: In this type of problems, we have to check separately for the points where the function splits. The function will be discontinuous at such particular points, wherever the graph breaks.