(no global maxima found)
Step-by-step explanation:
Find and classify the global extrema of the following function:
f(x) = 4 x^4 - 2 x^3 + x - 5
Hint: | Global extrema of f(x) can occur only at the critical points or the endpoints of the domain.
Find the critical points of f(x):
Compute the critical points of 4 x^4 - 2 x^3 + x - 5
Hint: | To find critical points, find where f'(x) is zero or where f'(x) does not exist. First, find the derivative of 4 x^4 - 2 x^3 + x - 5.
To find all critical points, first compute f'(x):
d/( dx)(4 x^4 - 2 x^3 + x - 5) = 16 x^3 - 6 x^2 + 1:
f'(x) = 16 x^3 - 6 x^2 + 1
Hint: | Find where f'(x) is zero by solving 16 x^3 - 6 x^2 + 1 = 0.
Solving 16 x^3 - 6 x^2 + 1 = 0 yields x≈-0.303504:
x = -0.303504
Hint: | Find where f'(x) = 16 x^3 - 6 x^2 + 1 does not exist.
f'(x) exists everywhere:
16 x^3 - 6 x^2 + 1 exists everywhere
Hint: | Collect results.
The only critical point of 4 x^4 - 2 x^3 + x - 5 is at x = -0.303504:
x = -0.303504
Hint: | Determine the endpoints of the domain of f(x).
The domain of 4 x^4 - 2 x^3 + x - 5 is R:
The endpoints of R are x = -∞ and ∞
Hint: | Evaluate f(x) at the critical points and at the endpoints of the domain, taking limits if necessary.
Evaluate 4 x^4 - 2 x^3 + x - 5 at x = -∞, -0.303504 and ∞:
The open endpoints of the domain are marked in gray
x | f(x)
-∞ | ∞
-0.303504 | -5.21365
∞ | ∞
Hint: | Determine the largest and smallest values that f achieves at these points.
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:
The open endpoints of the domain are marked in gray
x | f(x) | extrema type
-∞ | ∞ | global max
-0.303504 | -5.21365 | global min
∞ | ∞ | global max
Hint: | Finally, remove the endpoints of the domain where f(x) is not defined.
Remove the points x = -∞ and ∞ from the table
These cannot be global extrema, as the value of f(x) here is never achieved:
x | f(x) | extrema type
-0.303504 | -5.21365 | global min
Hint: | Summarize the results.
f(x) = 4 x^4 - 2 x^3 + x - 5 has one global minimum:
f(x) has a global minimum at x = -0.303504
