- Both functions are strictly increasing due to their positive slope \(m=2\).
- The range of \(g(x) = 2x+1\) on \((0, 5)\) is \((2(0)+1, 2(5)+1) = (1, 11)\).
- The range of \(h(x) = 2x+2\) on \((0, 5)\) is \((2(0)+2, 2(5)+2) = (2, 12)\).
- The range of \(h(x)\) is shifted upwards by 1 relative to \(g(x)\), with ranges of \(\boxed{(1, 11)}\) and \(\boxed{(2, 12)}\) respectively.
Explanation
- Analyzing Function Properties
We analyze the functions \(g(x) = 2x + 1\) and \(h(x) = 2x + 2\) within the specified domain \(0 < x < 5\). Both functions are linear with a positive slope of \(m = 2\). A linear function with a positive slope is strictly increasing, meaning that as \(x\) increases, the function value \(f(x)\) also increases. Given the open interval domain \((0, 5)\), the range of such a function is \((f(0), f(5))\). By observing the constant terms, \(g(x)\) has a constant term of 1 and \(h(x)\) has a constant term of 2. Since 2 is greater than 1, we can infer that for any value of \(x\), \(h(x) = g(x) + 1\). Therefore, the range of \(h(x)\) must be the same as the range of \(g(x)\) shifted upward by 1 unit.
- Calculating and Comparing Ranges
To determine the range of \(g(x) = 2x + 1\) on \((0, 5)\):
- At the lower boundary, as \(x \to 0^+\), \(g(x) \to 2(0) + 1 = 1\).
- At the upper boundary, as \(x \to 5^-\), \(g(x) \to 2(5) + 1 = 11\).
So the range of \(g(x)\) is \((1, 11)\).
To determine the range of \(h(x) = 2x + 2\) on \((0, 5)\):
- At the lower boundary, as \(x \to 0^+\), \(h(x) \to 2(0) + 2 = 2\).
- At the upper boundary, as \(x \to 5^-\), \(h(x) \to 2(5) + 2 = 12\).
So the range of \(h(x)\) is \((2, 12)\).
Comparing the two, we see that the range of \(h(x)\) is indeed \((1+1, 11+1) = (2, 12)\), confirming our previous observation that \(h(x)\) is shifted up by 1 unit compared to \(g(x)\).
- Graphical Description and Conclusion
The graph of these functions consists of two parallel line segments.
- The function \(g(x)\) is represented by a line segment starting from an open circle at \((0, 1)\) and ending at an open circle at \((5, 11)\).
- The function \(h(x)\) is represented by a line segment starting from an open circle at \((0, 2)\) and ending at an open circle at \((5, 12)\).
Because the slopes are the same, the lines are parallel. The vertical distance between the two segments is constant, specifically \(h(x) - g(x) = 2 - 1 = 1\). The range of \(g(x)\) is \(\boxed{(1, 11)}\) and the range of \(h(x)\) is \(\boxed{(2, 12)}\).
Examples
Understanding the vertical translation of linear functions is crucial in real-world scenarios such as logistics or finance. For example, if two delivery services charge a base fee plus a per-mile rate, their pricing can be modeled as linear functions. If Service A charges \$1 base plus \$2 per mile and Service B charges \$2 base plus \$2 per mile, the difference in the cost ranges is constant regardless of how many miles are driven. This type of analysis helps businesses and consumers compare pricing structures and understand how different base costs affect the total expenditure over a specific distance interval.