College Board AP Classroom Unit 5 Progress Check: MCQ Part B 10 11 12 Question 5 A Let gbe the function defined by g (z) = (z? – 1 + 1)e What in the absolute maximum value of g on the interval -4,1) O Type here to search
The Correct Answer and Explanation is :
The problem is asking for the absolute maximum value of the function ( g(z) = (z^2 – 1 + 1)e^z ) on the interval ( [-4, 1] ).
Step 1: Simplify the function
The given function is:
[
g(z) = (z^2 – 1 + 1)e^z
]
Simplifying the expression inside the parentheses:
[
g(z) = z^2 e^z
]
Step 2: Find the critical points
To find the absolute maximum, we first need to take the derivative of the function and find the critical points within the interval ( [-4, 1] ).
The derivative of ( g(z) = z^2 e^z ) is found using the product rule:
[
g'(z) = \frac{d}{dz}(z^2) \cdot e^z + z^2 \cdot \frac{d}{dz}(e^z)
]
[
g'(z) = 2z e^z + z^2 e^z
]
Factor out ( e^z ):
[
g'(z) = e^z(2z + z^2)
]
To find the critical points, set ( g'(z) = 0 ):
[
e^z(2z + z^2) = 0
]
Since ( e^z ) is never zero, we focus on solving:
[
2z + z^2 = 0
]
[
z(2 + z) = 0
]
This gives two critical points: ( z = 0 ) and ( z = -2 ).
Step 3: Evaluate the function at the critical points and endpoints
Now we evaluate ( g(z) ) at the critical points and the endpoints of the interval ( [-4, 1] ).
- At ( z = -4 ):
[
g(-4) = (-4)^2 e^{-4} = 16 e^{-4}
] - At ( z = -2 ):
[
g(-2) = (-2)^2 e^{-2} = 4 e^{-2}
] - At ( z = 0 ):
[
g(0) = 0^2 e^0 = 0
] - At ( z = 1 ):
[
g(1) = 1^2 e^1 = e
]
Step 4: Determine the maximum value
Now we compare the values:
- ( g(-4) = 16 e^{-4} \approx 0.0008 )
- ( g(-2) = 4 e^{-2} \approx 0.5413 )
- ( g(0) = 0 )
- ( g(1) = e \approx 2.7183 )
The absolute maximum value of ( g(z) ) on the interval ( [-4, 1] ) is ( g(1) = e ).
Final Answer:
The absolute maximum value of ( g(z) ) on the interval ( [-4, 1] ) is ( e \approx 2.7183 ).