Respuesta :
Answer:
Where 0 < x < 3
The location of the local minimum, is (2, 0)
The location of the local maximum is at (0, 16)
Step-by-step explanation:
The given function is f(x) = (x + 2)⁴
The range of the minimum = 0 < x < 3
At a local minimum/maximum values, we have;
[tex]f'(x) = \dfrac{(-x + 2)^4}{dx} = -4 \cdot (-x + 2)^3 = 0[/tex]
∴ (-x + 2)³ = 0
x = 2
[tex]f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx} = -12 \cdot (-x + 2)^2[/tex]
When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2
We have, f(2) = (-2 + 2)⁴ = 0
The location of the local minimum, is (2, 0)
Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3
When x = 0, -x + 2 = 0 + 2 = 2
Similarly, we have;
-x + 2 = 1, when x = 1
-x + 2 = 0, when x = 2
-x + 2 = -1, when x = 3
Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16
The location of the local maximum is at (0, 16).
