abs min and ABS MAX of 2cos(t)+sin(2t), (2014 version)

TL;DR

The video discusses how to find the maximum and minimum values of the function 2cos(t) + sin(2t) on the interval 0 to π/2.

Transcript

4.1 number 57 we are going to find our apps to maximum and the minimum value for the function 2 cosine t plus sine of 2t on the interval 0 comma pi over 2. to do this we have to get our critical numbers and the critical numbers means that we have to set our derivative equal to zero and that's one of the possibilities for this so we begin with our f... Read More

Key Insights

• 💉 To find the maximum and minimum values of a function, we need to find the critical numbers, which are the values of t where the derivative is equal to zero.
• 😫 By taking the derivative and setting it equal to zero, we can find the critical numbers for t.
• 🔢 Solving the equation -2sin(t) + cos(2t) = 0, we find the critical number t = π/6 within the interval of consideration.
• ✅ After checking the values at the critical number and the endpoints, we determine the absolute maximum to be 3√3/2 at t = π/6, and the absolute minimum to be 0 at t = π/2.
• 🕴️ The trig identities cos(2t) = 1 - 2sin^2(t) and the definition of sine were used to solve the quadratic equation.
• 🛀 Graphing the function shows a maximum and minimum occurring within the interval 0 to π/2.

Questions & Answers

Q: How do you find the critical numbers of a function?

To find the critical numbers, you need to set the derivative of the function equal to zero and solve for the variable t. In this case, we set -2sin(t) + cos(2t) = 0 and found t = π/6 as the only critical number in the given interval.

Q: How do you determine whether it is the maximum or minimum value?

To determine whether it is a maximum or minimum value, we need to check the values at the critical number and the endpoints. In this case, we found that the function had an absolute maximum value of 3√3/2 when t = π/6, and an absolute minimum value of 0 when t = π/2.

Q: How did you solve -2sin(t) + cos(2t) = 0?

We used the trig identity cos(2t) = 1 - 2sin^2(t) to rewrite the equation as -2sin(t) + 1 - 2sin^2(t) = 0. Simplifying further, we obtained -2sin^2(t) - 2sin(t) + 1 = 0. Factoring this quadratic equation, we got (2sin(t) - 1)(sin(t) + 1) = 0, which gave us the two equations 2sin(t) - 1 = 0 and sin(t) + 1 = 0 to solve for the critical numbers.

Q: How did you check the values at the critical number and endpoints?

To check the values, we substituted the critical number and endpoints (t = π/6, 0, and π/2) into the original function, 2cos(t) + sin(2t). When t = π/6, we found the value to be 3√3/2. When t = 0, the value was 2, and when t = π/2, the value was 0. These values helped us determine the absolute maximum and minimum.

Summary & Key Takeaways

• To find the maximum and minimum values of the function, we need to find the critical numbers, which are the values of t where the derivative is equal to zero.

• By taking the derivative of the function, we get -2sin(t) + 2cos(2t) = 0, which simplifies to -2sin(t) + cos(2t) = 0.

• Solving this equation, we find that the only critical number within the interval is t = π/6.

• After checking the values at the critical number and the endpoints of the interval, we determine that the absolute maximum value is 3√3/2 when t = π/6, and the absolute minimum value is 0 when t = π/2.