Find the Roots (Zeros) -xe^(-x)+e^(-x)

$- x e^{- x} + e^{- x}$

Step 1

Set $- x e^{- x} + e^{- x}$ equal to $0$ .

$- x e^{- x} + e^{- x} = 0$

Step 2

Solve for $x$ .

Factor $e^{- x}$ out of $- x e^{- x} + e^{- x}$ .

Factor $e^{- x}$ out of $- x e^{- x}$ .

$e^{- x} (- x) + e^{- x} = 0$

Multiply by $1$ .

$e^{- x} (- x) + e^{- x} \cdot 1 = 0$

Factor $e^{- x}$ out of $e^{- x} (- x) + e^{- x} \cdot 1$ .

$e^{- x} (- x + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$e^{- x} = 0$

$- x + 1 = 0$

Set $e^{- x}$ equal to $0$ and solve for $x$ .

Set $e^{- x}$ equal to $0$ .

$e^{- x} = 0$

Solve $e^{- x} = 0$ for $x$ .

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

$\ln (e^{- x}) = \ln (0)$

The equation cannot be solved because $\ln (0)$ is undefined.

Undefined

There is no solution for $e^{- x} = 0$

No solution

Set $- x + 1$ equal to $0$ and solve for $x$ .

Set $- x + 1$ equal to $0$ .

$- x + 1 = 0$

Solve $- x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

The final solution is all the values that make $e^{- x} (- x + 1) = 0$ true.

$x = 1$

Step 3

Find the Roots (Zeros) -xe^(-x)+e^(-x)

Related Questions

Support LCMGCF.com by disabling your adblocker.