Question

Let f(x)=\(\begin{Bmatrix}  x+1&\,\,\,-1\leq x\leq 0\\   -x&\,\,\,\,\,0<x\leq1    \end{Bmatrix}\)

• A. f(x) is discontinuous in [-1,1] and so has no maximum value or minimum value in [-1,1].
• B. f(x) is continuous in [-1,1] and so has maximum and minimum value
• C. f(x) is bounded in[-1,1] and doesn't attain maximum or minimum value
• D. f(x) is discontinuous in [-1,1] but still has the maximum and minimum value

Answer 1

The Correct Option is D

Step 1: Check for Continuity

We need to check the continuity at the point where the function definition changes, which is \(x = 0\).

The left-hand limit at \(x = 0\) is:

\(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x + 1) = 0 + 1 = 1\)

The right-hand limit at \(x = 0\) is:

\(\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (-x) = 0\)

The function value at \(x = 0\) is:

\(f(0) = 0 + 1 = 1\)

Since \(\lim_{x \to 0^-} f(x) = f(0) = 1 \neq \lim_{x \to 0^+} f(x) = 0\), the function is discontinuous at \(x = 0\).

Step 2: Determine Maximum and Minimum Values

Despite the discontinuity, let's analyze the function in the interval \([-1, 1]\).

For \(-1 \leq x \leq 0\), \(f(x) = x + 1\)

* At \(x = -1\), \(f(-1) = -1 + 1 = 0\) (minimum value)

* At \(x = 0\), \(f(0) = 0 + 1 = 1\)

For \(0 < x \leq 1\), \(f(x) = -x\)

* As \(x\) approaches \(0\) from the right, \(f(x)\) approaches \(0\).

* At \(x = 1\), \(f(1) = -1\) (this one isn't a candidate)

Let \(x_1=-1,x_2=0^-,x_3=0^+,x_4=1\) then \(f(x_1)=f(-1)=0,f(x_2)=f(0^-)=1,f(x_3)=f(0^+)=0,f(x_4)=f(1)=-1 \) We see there is a min at =-1 and max at 1

Conclusion:

\(f(x)\) is discontinuous in \([-1, 1]\), but it still has a maximum value (1) and a minimum value (-1) in this interval.

Therefore, the correct option is: \(f(x)\) is discontinuous in \([-1, 1]\) but still has the maximum and minimum value.