- Identify that the function \(h(t) = -t^2 + 10t - 16\) is positive between its roots \(2\) and \(8\), so \(2 < t < 8\).
- Find the derivative \(h'(t) = -2t + 10\) and determine it is positive (increasing) when \(t < 5\).
- Intersect the two conditions \(2 < t < 8\) and \(t < 5\) to find the shared interval.
- The final interval is \(2 < t < 5\), which corresponds to the last option.
- Final Answer: \(\boxed{2 < t < 5}\)
Explanation
- Analysis of the function \(h(t)\)
We are given the function \(h(t) = -t^2 + 10t - 16\). Our goal is to determine the interval of \(t\)-values for which the function is both positive (\(h(t) > 0\)) and increasing (\(h'(t) > 0\)). The function is a downward-opening parabola because the coefficient of \(t^2\) is negative (\(-1\)). A downward-opening parabola is positive between its two roots. We will find these roots and then find where the function is increasing using its derivative.
- Finding the interval where \(h(t) > 0\)
To find where \(h(t) > 0\), we first set the function equal to zero to find its roots: \(-t^2 + 10t - 16 = 0\). Multiplying by \(-1\) to simplify, we get \(t^2 - 10t + 16 = 0\). Factoring the quadratic, we find \((t - 2)(t - 8) = 0\), which gives the roots \(t = 2\) and \(t = 8\). Since the parabola opens downward, the function \(h(t)\) is positive for values between the two roots. Therefore, \(h(t) > 0\) when \(2 < t < 8\).
- Finding the interval where \(h(t)\) is increasing
The function is increasing when its derivative \(h'(t)\) is greater than zero. The derivative is calculated as \(h'(t) = \frac{d}{dt}(-t^2 + 10t - 16) = -2t + 10\). To find where it is increasing, we solve the inequality \(-2t + 10 > 0\). Subtracting 10 from both sides gives \(-2t > -10\), and dividing by \(-2\) (which reverses the inequality sign) results in \(t < 5\). Thus, the function is increasing for all \(t < 5\).
- Determining the overlapping interval
We now look for the intersection of the two intervals: \(2 < t < 8\) (where the function is positive) and \(t < 5\) (where the function is increasing). The common values that satisfy both conditions are \(2 < t < 5\). We conclude that the function is both positive and increasing on the interval \(2 < t < 5\).
Examples
A real-life scenario where this concept is used is in optimizing production or profit cycles. For instance, if \(h(t)\) represents the profit of a business over time \(t\) (in months), understanding the interval \(2 < t < 5\) helps the manager know when the business is both making a profit and experiencing growth in that profit. This allows them to focus resources during the most efficient growth phase to maximize their results before the market conditions potentially change after month 5.