Question

For what value of \(\lambda\), is the function \[f(x)=\begin{cases}\lambda(x^2-2x), & x\le0 \\ 4x+1, & x>0\end{cases}\] continuous at \(x=0\)?

Hint:

Find the left hand limit and right hand limit of f(x) at x = 0 and check whether they can be made equal.

Answer 1

Step 1: Understanding the Concept:
A function is continuous at a point when the left hand limit, the right hand limit and the value of the function at that point are all equal.
Here f is defined by two separate rules on either side of x = 0, so each of these three quantities must be found using the correct rule.

Step 2: Find the Left Hand Limit:
For x less than or equal to 0, the rule is \(f(x)=\lambda(x^2-2x)\).
As x approaches 0 from the left, both \(x^2\) and \(x\) tend to 0, so the bracket tends to 0 for any value of \(\lambda\).
\[ \lim_{x\to0^-}f(x)=\lambda(0^2-2\cdot0)=0 \]

Step 3: Find the Right Hand Limit and f(0):
For x greater than 0, the rule is \(f(x)=4x+1\), so the right hand limit is obtained by putting x = 0 in this rule.
Since 0 satisfies \(x\le0\), the value \(f(0)\) is taken from the first rule and equals 0 as well.
\[ \lim_{x\to0^+}f(x)=4(0)+1=1, \qquad f(0)=\lambda(0)=0 \]

Step 4: Compare the Three Values:
The left hand limit is 0 and the right hand limit is 1, and neither of these values contains \(\lambda\), so they are fixed numbers.
For continuity we need \(0=1\), which is false for every real number, so no choice of \(\lambda\) can satisfy the condition.

Final Answer:
The one sided limits 0 and 1 can never be made equal, so continuity at x = 0 is impossible for any lambda.
\[ \boxed{\text{No real value of } \lambda \text{ makes } f(x) \text{ continuous at } x=0} \]