Find the points at which y=f(x)=x11−7x has a global maximum and minimum on the interval 0≤x≤2.5. Round your answers to two decimal places.
Global max:
Global min:
Global max: At x = 0.00, because it's the lowest possible value for x. It's so low, it's rock bottom!
Global min: At x = 2.50, because it's close to the end of the interval. It's like finding the last piece of chocolate cake at a party. Cherish it!
To find the points at which the function has a global maximum and minimum, we need to first find the critical points of the function.
1. Find the derivative of the function:
f'(x) = 11x^10 - 7
2. Set the derivative equal to zero and solve for x:
11x^10 - 7 = 0
To solve this equation, we can't solve it algebraically, so we can use numerical methods such as the Newton-Raphson method or graphing calculator.
Using a graphing calculator, we find that x ≈ 0.717 is a critical point.
Now, we need to check the endpoints and critical points to determine the global maximum and minimum.
3. Evaluate the function at the endpoints and the critical point:
f(0) = (0)^11 - 7(0) = 0
f(2.5) = (2.5)^11 - 7(2.5) ≈ 7.593
f(0.717) ≈ -4.106
4. Compare the values of f(x) at the endpoints and the critical point.
The global maximum is f(2.5) ≈ 7.593, which occurs at x = 2.5.
The global minimum is f(0.717) ≈ -4.106, which occurs at x ≈ 0.717.
Therefore, the points at which the function has a global maximum and minimum on the interval 0 ≤ x ≤ 2.5 are:
Global max: (2.5, 7.593)
Global min: (0.717, -4.106)
To find the points at which a function has a global maximum and minimum on a specific interval, you can follow these steps:
1. Find the critical points of the function on the given interval by taking the derivative of the function and setting it equal to zero.
2. Evaluate the function at the critical points and the endpoints of the interval.
3. Compare the values obtained in step 2 to determine the global maximum and minimum.
Let's apply these steps to the given function, y=f(x)=x^11−7x, on the interval 0≤x≤2.5.
1. Find the critical points:
Taking the derivative of the function, we get:
f'(x) = 11x^10 - 7
Setting f'(x) equal to zero and solving for x, we get:
11x^10 - 7 = 0
To solve this equation, we can factor out the common factor:
x^10 = 7/11
Taking the 10th root of both sides, we have:
x = (7/11)^(1/10)
This gives us one critical point.
2. Evaluate the function:
Evaluate f(x) at the critical point (x = (7/11)^(1/10)) and the endpoints of the interval (x = 0 and x = 2.5).
f(0) = (0)^11 - 7(0) = 0 - 0 = 0
f((7/11)^(1/10)) = ((7/11)^(1/10))^11 - 7((7/11)^(1/10))
Calculating this value depends on the specific value of (7/11)^(1/10).
f(2.5) = (2.5)^11 - 7(2.5)
3. Compare the values:
Compare the values obtained in the previous step to determine the global maximum and minimum. Select the highest and lowest values among the critical point, endpoint values, and evaluate f(x) at those points to determine the precise coordinates.
Once you have calculated the values for f(0), f((7/11)^(1/10)), and f(2.5), you can determine the global maximum and minimum by comparing these values.
Round your answers to two decimal places to get the precise coordinates of the global maximum and minimum on the interval 0≤x≤2.5.
y = x^11 - 7x
y' = 11x^10 - 7
so local extrema at x = ±(7/11)^(1/10) ≈ ±0.96
Our domain is [0,2.5] so we just have to worry about _0.96
Since y" > 0 there, that is a local minimum.
y(0) = 0
y(2.5) > 0, so
y(0.96) is the global min
y(2.5) is the global max