Find the Critical Points xe^(-2x)

Problem

ƒ(x)=x*e^(−2*x)

Solution

1. Identify the function ƒ(x)=x*e^(−2*x) and recall that critical points occur where the first derivative ƒ(x)^′ is equal to zero or is undefined.

2. Apply the product rule for differentiation, which states (d(u)*v)/d(x)=ud(v)/d(x)+vd(u)/d(x) where u=x and v=e^(−2*x)

3. Differentiate the components: d(x)/d(x)=1 and d(e^(−2*x))/d(x)=−2*e^(−2*x) using the chain rule.

4. Combine the results to find the derivative:

ƒ(x)^′=x*(−2*e^(−2*x))+e(1)^(−2*x)

5. Factor out the common term e^(−2*x) to simplify the expression:

ƒ(x)^′=e^(−2*x)*(1−2*x)

6. Set the derivative to zero to find the critical points:

e^(−2*x)*(1−2*x)=0

7. Solve for x by noting that e^(−2*x) is never zero for any real x so we must solve $1 - 2x = 0$.

1=2*x

x=1/2

Final Answer

x=1/2

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