Original Equation: f(x) = e^(-x) - e^(-2x)
Did I take the first derivative correctly?
I seem to get undefined when I isolate for x to determine the extreme values?
f'(x) = -e/ x + 2e/2x
f'(x) = (-2ex + 2ex) / 2x^2
What does e/x represent? e divided by x or e^x ? In either case, I don't agree with your answer.
To determine if you took the first derivative correctly, let's go through the process step by step.
The original equation is f(x) = e^(-x) - e^(-2x). To find the first derivative, we will use the rules of differentiation.
Step 1: Apply the power rule for derivatives:
If f(x) = ax^n, then f'(x) = nax^(n-1).
Using this rule, the derivative of e^(-x) is (-1)e^(-x) = -e^(-x), and the derivative of e^(-2x) is (-2)e^(-2x) = -2e^(-2x).
So, the equation for the first derivative f'(x) becomes:
f'(x) = -e^(-x) - (-2e^(-2x))
= -e^(-x) + 2e^(-2x)
Now, let's address your concern about determining the extreme values.
To find the extreme values, we need to solve for x when f'(x) = 0. In other words, we are looking for the values of x that make the first derivative equal to zero.
Setting f'(x) = 0, we have:
-e^(-x) + 2e^(-2x) = 0
Now, we need to isolate x to determine its values. Here's how to proceed:
1. Multiply the entire equation by e^(2x) to get rid of the denominators:
-e^(-x)e^(2x) + 2e^(-2x)e^(2x) = 0
2. Simplify the equation:
-e^(x) + 2 = 0
3. Add e^(x) to both sides:
2 = e^(x)
4. Take the natural logarithm (ln) of both sides to solve for x:
ln(2) = x
Therefore, the value of x that satisfies f'(x) = 0 is x = ln(2).
In summary, you correctly obtained the first derivative as f'(x) = -e^(-x) + 2e^(-2x). However, when solving for the extreme values, it seems like there might have been an error in isolating x. Make sure to follow the steps described above to obtain the correct values of x.