Find the points at which y = f(x) = x8−13x has a global maximum and minimum on the interval 0 ≤ x ≤ 3.7. Round your answers to two decimal places.

Global Maximum:
(x,y) = (,)
Global Minimum:
(x,y) = (,)

I just did the previous question for you, do this one the same way,

i cant figure out the global minimum

I will assume your function is

f(x) = x^8 - 13x
f(0) = 0
f(3.7) = 3.7^8 - 13(3.7) = appr. 35077

also f'(x) = 8x^7 - 13 = 0 for local max/min
x^7 = 13/8
x = 1.0718

f(1.0718) = approx -12.2

draw your conclusion following your course's definition of "global minimum"

To find the global maximum and minimum of the function y = f(x) = x^8 - 13x on the interval 0 ≤ x ≤ 3.7, we can follow these steps:

1. Find the critical points by taking the first derivative of f(x) and setting it equal to zero.
f'(x) = 8x^7 - 13 = 0

Solving this equation will give us the critical points.

2. Check the endpoints of the interval.
Evaluate y at x = 0 and x = 3.7 to see if they are higher or lower than the critical points.

3. Compare the values of y at the critical points and endpoints to determine the global maximum and minimum.

Let's go through these steps:

1. Find the critical points:
f'(x) = 8x^7 - 13 = 0

To solve this equation, we need to isolate x:
8x^7 = 13
x^7 = 13/8
x = (13/8)^(1/7)

Using a calculator or mathematical software, we find that x ≈ 1.4081 is the critical point.

2. Check the endpoints:
Evaluate f(0): f(0) = (0^8) - 13(0) = 0
Evaluate f(3.7): f(3.7) = (3.7^8) - 13(3.7) ≈ 9596.11

We see that f(0) = 0 and f(3.7) ≈ 9596.11.

3. Compare the values:
Comparing the critical point x ≈ 1.4081 to the endpoints, we can determine the global maximum and minimum.

Since f(0) = 0 is lower than the function value at the critical point, it cannot be the global maximum.

Therefore, the global maximum occurs at the critical point x ≈ 1.4081.
Plugging this value into the function, we can determine the y-coordinate:

y = f(x) = (1.4081^8) - 13(1.4081) ≈ -14.48

Rounding the answers to two decimal places:

Global Maximum: (x, y) ≈ (1.41, -14.48)

Since f(3.7) ≈ 9596.11 is higher than the function value at the critical point, it cannot be the global minimum.

Therefore, the global minimum occurs at the critical point x ≈ 1.4081.
Plugging this value into the function, we can determine the y-coordinate:

y = f(x) = (1.4081^8) - 13(1.4081) ≈ -14.48

Rounding the answers to two decimal places:

Global Minimum: (x, y) ≈ (1.41, -14.48)

Thus, the global maximum and minimum of the function on the given interval are approximately (1.41, -14.48).