Find the Domain and Range f(x)=(x+1)e^x

Problem

ƒ(x)=(x+1)*ex

Solution

1. Identify the domain of the function. The expression (x+1)*ex consists of a polynomial and an exponential function, both of which are defined for all real numbers.

2. Determine the domain. Since there are no denominators that could be zero and no square roots of negative numbers, the domain is all real numbers.

D=(−∞,∞)

3. Find the first derivative to locate critical points for the range. Use the product rule: (d(u)*v)/d(x)=ud(v)/d(x)+vd(u)/d(x)

d(ƒ(x))/d(x)=(x+1)*ex+e(1)1

d(ƒ(x))/d(x)=(x+2)*ex

4. Solve for critical points by setting the derivative to zero. Since ex is never zero, we solve x+2=0

x=−2

5. Evaluate the function at the critical point to find the absolute minimum.

ƒ*(−2)=(−2+1)*e(−2)

ƒ*(−2)=−e(−2)

ƒ*(−2)=−1/(e2)

6. Analyze the end behavior of the function. As x→∞ ƒ(x)→∞ As x→−∞ ƒ(x)→0 because the exponential decay dominates the linear growth.

7. Determine the range. Since the absolute minimum is −1/(e2) and the function increases to infinity, the range includes all values from the minimum upward.

R=[−1/(e2),∞)

Final Answer

Domain: *(−∞,∞), Range: *[−1/(e2),∞)

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