Question

Consider the following function K(t)= 10t/t²+25 a. On what time intervals is the function K(t) increasing? b. On what intervals is it decreasing?

c. Determine any local extrema and their locations. Please complete the question to include part (c).

Answer 1

The function K(t) is increasing on the intervals (-∞, -5) and (5, ∞), and decreasing on the interval (-5, 5). It has a local maximum at t = -5 and a local minimum at t = 5.

Step-by-step explanation:

To determine on what time intervals the function K(t) = 10t / (t² + 25) is increasing or decreasing, and to find any local extrema, we must consider its derivative.

Step 1: Find the derivative of K(t). Using quotient rule, we get:

K'(t) = (t² + 25)(10) - (10t)(2t) / (t² + 25)²

K'(t) = (10t² + 250 - 20t²) / (t² + 25)²

K'(t) = (-10t² + 250) / (t² + 25)²

Step 2: Set the derivative equal to zero to find critical points:

-10t² + 250 = 0

t² = 25

t = ±5

Step 3: Use the first derivative test to determine if these points are maxima or minima. We test intervals around -5 and 5 using the first derivative:

• For t < -5, K'(t) is positive; thus K(t) is increasing.
• For -5 < t < 5, K'(t) is negative; thus K(t) is decreasing.
• For t > 5, K'(t) is positive; thus K(t) is increasing.

Local extrema: The function has a local maximum at t = -5 and a local minimum at t = 5. To find the values of the local extrema, plug the t values into K(t):

Local maximum at: K(-5) = 10(-5) / ((-5)²+25)

Local minimum at: K(5) = 10(5) / (5²+25)

Conclusion: The function K(t) is increasing on the intervals (-∞, -5) and (5, ∞), and decreasing on the interval (-5, 5). The local maximum is at t = -5 and the local minimum at t = 5.