2. Determine the minimum value of the function f(x)= e^(-x) - 2e^x on the interval -1≤x≤2.
Answer is minimum value 7.12?
To determine the minimum value of the function f(x) = e^(-x) - 2e^x on the interval -1 ≤ x ≤ 2, we can follow these steps:
Step 1: Find the critical points by taking the derivative of the function.
The derivative of f(x) is:
f'(x) = -e^(-x) - 2e^x
Step 2: Set f'(x) = 0 and solve for x to find the critical points.
-e^(-x) - 2e^x = 0
Rearranging the equation gives:
e^(-x) = -2e^x
Taking the natural logarithm (ln) of both sides, we get:
-ln(e^x) = ln(-2) + x
Simplifying further, we have:
-x = ln(-2) + x
Combining like terms, we get:
2x = -ln(2)
Dividing by 2, we find:
x = -ln(2)/2
Step 3: Check the endpoints of the interval.
We need to evaluate the function at the endpoints of the interval (-1 ≤ x ≤ 2) to see if any of them produce a minimum value.
- When x = -1:
f(-1) = e^(-(-1)) - 2e^(-1)
= e + 2e^(-1)
- When x = 2:
f(2) = e^(-2) - 2e^2
Step 4: Compare the values of f(x) at the critical points and endpoints.
Now we can compare the values of f(x) at the critical point x = -ln(2)/2 and the endpoints x = -1 and x = 2, and determine the minimum value.
f(-1) ≈ 7.39
f(-ln(2)/2) ≈ 7.12
f(2) ≈ -13.65
Based on these calculations, we can see that the minimum value of the function occurs at x ≈ -ln(2)/2, and the minimum value is approximately around 7.12.