Problem #4: Consider the following function.
[tex]f(x)=7 e^{x^2-2 x}, \quad 0 \leq x \leq 2[/tex]
(a) Find the smallest value of [tex]f(x)[/tex] on the given interval.
(b) Find the largest value of [tex]f(x)[/tex] on the given interval.
Answer (2)
Find the derivative of the function f ( x ) = 7 e x 2 − 2 x , which is f ′ ( x ) = 7 e x 2 − 2 x ( 2 x − 2 ) .
Set the derivative equal to zero to find critical points: 2 x − 2 = 0 , so x = 1 .
Evaluate the function at the critical point and endpoints: f ( 0 ) = 7 , f ( 1 ) = e 7 , f ( 2 ) = 7 .
Determine the smallest and largest values: smallest is e 7 and largest is 7 , so the final answers are e 7 and 7 .
Explanation
Problem Analysis We are given the function f ( x ) = 7 e x 2 − 2 x and asked to find its smallest and largest values on the interval 0 l e q x ≤ 2 . To do this, we need to find the critical points of the function and evaluate the function at these points and at the endpoints of the interval.
Finding Critical Points First, we find the derivative of f ( x ) using the chain rule: f ′ ( x ) = 7 e x 2 − 2 x ⋅ ( 2 x − 2 ) To find the critical points, we set f ′ ( x ) = 0 : 7 e x 2 − 2 x ( 2 x − 2 ) = 0 Since 7 e x 2 − 2 x is always positive, we only need to solve 2 x − 2 = 0 , which gives us x = 1 .
Evaluating the Function Now we evaluate f ( x ) at the critical point x = 1 and at the endpoints x = 0 and x = 2 : f ( 0 ) = 7 e 0 2 − 2 ( 0 ) = 7 e 0 = 7 f ( 1 ) = 7 e 1 2 − 2 ( 1 ) = 7 e − 1 = e 7 f ( 2 ) = 7 e 2 2 − 2 ( 2 ) = 7 e 0 = 7
Determining Minimum and Maximum Comparing the values, we have f ( 0 ) = 7 , f ( 1 ) = e 7 , and f ( 2 ) = 7 . Since e ≈ 2.718 , we know that e 7 < 7 . Therefore, the smallest value of f ( x ) on the interval is e 7 , and the largest value is 7 .
Examples
Consider a scenario where you're analyzing the temperature of a chemical reaction over time. The function f ( x ) = 7 e x 2 − 2 x could represent the temperature of the reaction at time x , where x is between 0 and 2 hours. Finding the minimum and maximum values of this function helps you determine the lowest and highest temperatures the reaction reaches, which is crucial for safety and efficiency. This type of analysis is essential in chemical engineering to ensure reactions are stable and controlled.
The smallest value of the function f ( x ) = 7 e x 2 − 2 x on the interval [ 0 , 2 ] is e 7 . The largest value is 7 at the endpoints. Thus, the final results are e 7 and 7 .
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