Final answer:
The absolute minimum and absolute maximum of the function g(x) = eˣ sin(x) over the closed interval [0, π] can be found by evaluating the function at the critical points and the endpoints of the interval. The function has an absolute minimum at (0, 0) and an absolute maximum at (3π/4, e⁽³π/⁴⁾).
Step-by-step explanation:
The absolute minimum and absolute maximum of the function g(x) = eˣ sin(x) over the closed interval [0, π] can be found by evaluating the function at the critical points and the endpoints of the interval.
To find the critical points, we need to take the derivative of g(x):
g'(x) = eˣ sin(x) + eˣ cos(x)
Setting g'(x) = 0, we obtain:
eˣ sin(x) + eˣ cos(x) = 0
Factoring out eˣ, we have:
eˣ (sin(x) + cos(x)) = 0
Therefore, eˣ = 0 or sin(x) + cos(x) = 0.
Since eˣ is always positive, we can ignore eˣ = 0. So, we need to solve sin(x) + cos(x) = 0.
By using trigonometric identities, we can rewrite sin(x) + cos(x) = 0 as:
√2 sin(x + π/4) = 0
This equation is satisfied when x + π/4 = 0 or x + π/4 = π.
Solving for x, we find two critical points:
x = -π/4 and x = 3π/4.
We also need to evaluate g(x) at the endpoints of the interval [0, π].
g(0) = e⁰ sin(0) = 0
g(π) = eˣ sin(π) = 0
So, the function g(x) has an absolute minimum at (0, 0) and an absolute maximum at (3π/4, e⁽³π/⁴⁾).
Therefore, the correct answer is A) Absolute minimum at (0, 0) and absolute maximum at (π/2, e⁽π/²⁾).