Answer :
To analyze the function \( k(x) \) on the interval \((-\infty, 1)\), we need to focus on the piece of the function defined for \( x < 1 \). This part of the function is given as:
[tex]\[ k(x) = \frac{2}{x} \][/tex]
1. General Shape:
- The function \( k(x) = \frac{2}{x} \) is a hyperbola. A hyperbola typically has two asymptotes and is symmetric about these asymptotes.
2. Asymptotic Behavior:
- As \( x \) approaches \( 0 \) from the left (negative side), \( \frac{2}{x} \) becomes very large negative.
- As \( x \) approaches \( 0 \) from the right (positive side), \( \frac{2}{x} \) becomes very large positive.
3. Behavior at Endpoints:
- As \( x \) approaches \(-\infty\), \( \frac{2}{x} \) approaches \( 0 \) from the negative side.
- As \( x \) approaches \( 1 \) from the left, \( \frac{2}{x} \) approaches \( 2 \) because \( \frac{2}{1} = 2 \).
4. Decreasing/Increasing Intervals:
- On the interval \((-\infty, 0)\), the function \( \frac{2}{x} \) is negative and as the value of \( x \) becomes more negative, \( \frac{2}{x} \) approaches \( 0 \), meaning the function is decreasing.
- On the interval \((0, 1)\), the function \( \frac{2}{x} \) is positive and as the value of \( x \) approaches \( 1 \), \( \frac{2}{x} \) approaches \( 2 \), meaning the function is increasing.
From this, we deduce that:
- The general shape of the function \( k(x) = \frac{2}{x} \) on the interval \((-\infty, 1)\) is a hyperbola.
- The function \( k(x) \) is decreasing on the interval \((-\infty, 0)\) and increasing on the interval \((0, 1)\).
Thus, the answer is:
- Shape: Hyperbola
- Direction: Decreasing on [tex]\((-\infty, 0)\)[/tex] and increasing on [tex]\((0, 1)\)[/tex]
[tex]\[ k(x) = \frac{2}{x} \][/tex]
1. General Shape:
- The function \( k(x) = \frac{2}{x} \) is a hyperbola. A hyperbola typically has two asymptotes and is symmetric about these asymptotes.
2. Asymptotic Behavior:
- As \( x \) approaches \( 0 \) from the left (negative side), \( \frac{2}{x} \) becomes very large negative.
- As \( x \) approaches \( 0 \) from the right (positive side), \( \frac{2}{x} \) becomes very large positive.
3. Behavior at Endpoints:
- As \( x \) approaches \(-\infty\), \( \frac{2}{x} \) approaches \( 0 \) from the negative side.
- As \( x \) approaches \( 1 \) from the left, \( \frac{2}{x} \) approaches \( 2 \) because \( \frac{2}{1} = 2 \).
4. Decreasing/Increasing Intervals:
- On the interval \((-\infty, 0)\), the function \( \frac{2}{x} \) is negative and as the value of \( x \) becomes more negative, \( \frac{2}{x} \) approaches \( 0 \), meaning the function is decreasing.
- On the interval \((0, 1)\), the function \( \frac{2}{x} \) is positive and as the value of \( x \) approaches \( 1 \), \( \frac{2}{x} \) approaches \( 2 \), meaning the function is increasing.
From this, we deduce that:
- The general shape of the function \( k(x) = \frac{2}{x} \) on the interval \((-\infty, 1)\) is a hyperbola.
- The function \( k(x) \) is decreasing on the interval \((-\infty, 0)\) and increasing on the interval \((0, 1)\).
Thus, the answer is:
- Shape: Hyperbola
- Direction: Decreasing on [tex]\((-\infty, 0)\)[/tex] and increasing on [tex]\((0, 1)\)[/tex]
