Calculus Examples

$f (x) = x^{2} e^{- x}$

Step 1

Find the first derivative of the function.

Tap for more steps...

Step 1.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{2}$ and $g (x) = e^{- x}$ .

$x^{2} \frac{d}{d x} [e^{- x}] + e^{- x} \frac{d}{d x} [x^{2}]$

Step 1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x$ .

Tap for more steps...

Step 1.2.1

To apply the Chain Rule, set $u$ as $- x$ .

$x^{2} (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 1.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$x^{2} (e^{u} \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 1.2.3

Replace all occurrences of $u$ with $- x$ .

$x^{2} (e^{- x} \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 1.3

Differentiate.

Tap for more steps...

Step 1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$x^{2} (e^{- x} (- \frac{d}{d x} [x])) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$x^{2} (e^{- x} (- 1 \cdot 1)) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 1.3.3

Simplify the expression.

Tap for more steps...

Step 1.3.3.1

Multiply $- 1$ by $1$ .

$x^{2} (e^{- x} \cdot - 1) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 1.3.3.2

Move $- 1$ to the left of $e^{- x}$ .

$x^{2} (- 1 \cdot e^{- x}) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 1.3.3.3

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$x^{2} (- e^{- x}) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 1.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$x^{2} (- e^{- x}) + e^{- x} (2 x)$

Step 1.4

Simplify.

Tap for more steps...

Step 1.4.1

Reorder terms.

$- e^{- x} x^{2} + 2 e^{- x} x$

Step 1.4.2

Reorder factors in $- e^{- x} x^{2} + 2 e^{- x} x$ .

$- x^{2} e^{- x} + 2 x e^{- x}$

Step 2

Find the second derivative of the function.

Tap for more steps...

Step 2.1

By the Sum Rule, the derivative of $- x^{2} e^{- x} + 2 x e^{- x}$ with respect to $x$ is $\frac{d}{d x} [- x^{2} e^{- x}] + \frac{d}{d x} [2 x e^{- x}]$ .

$f'' (x) = \frac{d}{d x} (- x^{2} e^{- x}) + \frac{d}{d x} (2 x e^{- x})$

Step 2.2

Evaluate $\frac{d}{d x} [- x^{2} e^{- x}]$ .

Tap for more steps...

Step 2.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2} e^{- x}$ with respect to $x$ is $- \frac{d}{d x} [x^{2} e^{- x}]$ .

$f'' (x) = - \frac{d}{d x} (x^{2} e^{- x}) + \frac{d}{d x} (2 x e^{- x})$

Step 2.2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{2}$ and $g (x) = e^{- x}$ .

$f'' (x) = - (x^{2} \frac{d}{d x} (e^{- x}) + e^{- x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{- x})$

Step 2.2.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x$ .

Tap for more steps...

Step 2.2.3.1

To apply the Chain Rule, set $u_{1}$ as $- x$ .

$f'' (x) = - (x^{2} (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{- x})$

Step 2.2.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = - (x^{2} (e^{u_{1}} \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{- x})$

Step 2.2.3.3

Replace all occurrences of $u_{1}$ with $- x$ .

$f'' (x) = - (x^{2} (e^{- x} \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{- x})$

Step 2.2.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$f'' (x) = - (x^{2} (e^{- x} (- \frac{d}{d x} x)) + e^{- x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{- x})$

Step 2.2.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = - (x^{2} (e^{- x} (- 1 \cdot 1)) + e^{- x} \frac{d}{d x} (x^{2})) + \frac{d}{d x} (2 x e^{- x})$

Step 2.2.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$f'' (x) = - (x^{2} (e^{- x} (- 1 \cdot 1)) + e^{- x} (2 x)) + \frac{d}{d x} (2 x e^{- x})$

Step 2.2.7

Multiply $- 1$ by $1$ .

$f'' (x) = - (x^{2} (e^{- x} \cdot - 1) + e^{- x} (2 x)) + \frac{d}{d x} (2 x e^{- x})$

Step 2.2.8

Move $- 1$ to the left of $e^{- x}$ .

$f'' (x) = - (x^{2} (- 1 \cdot e^{- x}) + e^{- x} (2 x)) + \frac{d}{d x} (2 x e^{- x})$

Step 2.2.9

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + \frac{d}{d x} (2 x e^{- x})$

Step 2.3

Evaluate $\frac{d}{d x} [2 x e^{- x}]$ .

Tap for more steps...

Step 2.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x e^{- x}$ with respect to $x$ is $2 \frac{d}{d x} [x e^{- x}]$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 2 \frac{d}{d x} (x e^{- x})$

Step 2.3.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{- x}$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 2 (x \frac{d}{d x} (e^{- x}) + e^{- x} \frac{d}{d x} (x))$

Step 2.3.3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x$ .

Tap for more steps...

Step 2.3.3.1

To apply the Chain Rule, set $u_{2}$ as $- x$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 2 (x (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x))$

Step 2.3.3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 2 (x (e^{u_{2}} \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x))$

Step 2.3.3.3

Replace all occurrences of $u_{2}$ with $- x$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 2 (x (e^{- x} \frac{d}{d x} (- x)) + e^{- x} \frac{d}{d x} (x))$

Step 2.3.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 2 (x (e^{- x} (- \frac{d}{d x} x)) + e^{- x} \frac{d}{d x} (x))$

Step 2.3.5

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 2 (x (e^{- x} (- 1 \cdot 1)) + e^{- x} \frac{d}{d x} (x))$

Step 2.3.6

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 2 (x (e^{- x} (- 1 \cdot 1)) + e^{- x} \cdot 1)$

Step 2.3.7

Multiply $- 1$ by $1$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 2 (x (e^{- x} \cdot - 1) + e^{- x} \cdot 1)$

Step 2.3.8

Move $- 1$ to the left of $e^{- x}$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 2 (x (- 1 \cdot e^{- x}) + e^{- x} \cdot 1)$

Step 2.3.9

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 2 (x (- e^{- x}) + e^{- x} \cdot 1)$

Step 2.3.10

Multiply $e^{- x}$ by $1$ .

$f'' (x) = - (x^{2} (- e^{- x}) + e^{- x} (2 x)) + 2 (x (- e^{- x}) + e^{- x})$

Step 2.4

Simplify.

Tap for more steps...

Step 2.4.1

Apply the distributive property.

$f'' (x) = - (x^{2} (- e^{- x})) - (e^{- x} (2 x)) + 2 (x (- e^{- x}) + e^{- x})$

Step 2.4.2

Apply the distributive property.

$f'' (x) = - (x^{2} (- e^{- x})) - (e^{- x} (2 x)) + 2 (x (- e^{- x})) + 2 e^{- x}$

Step 2.4.3

Combine terms.

Tap for more steps...

Step 2.4.3.1

Multiply $- 1$ by $- 1$ .

$f'' (x) = 1 (x^{2} (e^{- x})) - (e^{- x} (2 x)) + 2 (x (- e^{- x})) + 2 e^{- x}$

Step 2.4.3.2

Multiply $x^{2}$ by $1$ .

$f'' (x) = x^{2} e^{- x} - (e^{- x} (2 x)) + 2 (x (- e^{- x})) + 2 e^{- x}$

Step 2.4.3.3

Multiply $2$ by $- 1$ .

$f'' (x) = x^{2} e^{- x} - 2 (e^{- x} (x)) + 2 (x (- e^{- x})) + 2 e^{- x}$

Step 2.4.3.4

Multiply $- 1$ by $2$ .

$f'' (x) = x^{2} e^{- x} - 2 e^{- x} x - 2 (x (e^{- x})) + 2 e^{- x}$

Step 2.4.3.5

Subtract $2 x e^{- x}$ from $- 2 e^{- x} x$ .

Tap for more steps...

Step 2.4.3.5.1

Move $e^{- x}$ .

$f'' (x) = x^{2} e^{- x} - 2 x e^{- x} - 2 x e^{- x} + 2 e^{- x}$

Step 2.4.3.5.2

Subtract $2 x e^{- x}$ from $- 2 x e^{- x}$ .

$f'' (x) = x^{2} e^{- x} - 4 x e^{- x} + 2 e^{- x}$

Step 2.4.4

Reorder terms.

$f'' (x) = e^{- x} x^{2} - 4 e^{- x} x + 2 e^{- x}$

Step 2.4.5

Reorder factors in $e^{- x} x^{2} - 4 e^{- x} x + 2 e^{- x}$ .

$f'' (x) = x^{2} e^{- x} - 4 x e^{- x} + 2 e^{- x}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$- x^{2} e^{- x} + 2 x e^{- x} = 0$

Step 4

Find the first derivative.

Tap for more steps...

Step 4.1

Find the first derivative.

Tap for more steps...

Step 4.1.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{2}$ and $g (x) = e^{- x}$ .

$x^{2} \frac{d}{d x} [e^{- x}] + e^{- x} \frac{d}{d x} [x^{2}]$

Step 4.1.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - x$ .

Tap for more steps...

Step 4.1.2.1

To apply the Chain Rule, set $u$ as $- x$ .

$x^{2} (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 4.1.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u} [a^{u}]$ is $a^{u} \ln (a)$ where $a$ = $e$ .

$x^{2} (e^{u} \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 4.1.2.3

Replace all occurrences of $u$ with $- x$ .

$x^{2} (e^{- x} \frac{d}{d x} [- x]) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 4.1.3

Differentiate.

Tap for more steps...

Step 4.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$x^{2} (e^{- x} (- \frac{d}{d x} [x])) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 4.1.3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 1$ .

$x^{2} (e^{- x} (- 1 \cdot 1)) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 4.1.3.3

Simplify the expression.

Tap for more steps...

Step 4.1.3.3.1

Multiply $- 1$ by $1$ .

$x^{2} (e^{- x} \cdot - 1) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 4.1.3.3.2

Move $- 1$ to the left of $e^{- x}$ .

$x^{2} (- 1 \cdot e^{- x}) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 4.1.3.3.3

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$x^{2} (- e^{- x}) + e^{- x} \frac{d}{d x} [x^{2}]$

Step 4.1.3.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 2$ .

$x^{2} (- e^{- x}) + e^{- x} (2 x)$

Step 4.1.4

Simplify.

Tap for more steps...

Step 4.1.4.1

Reorder terms.

$- e^{- x} x^{2} + 2 e^{- x} x$

Step 4.1.4.2

Reorder factors in $- e^{- x} x^{2} + 2 e^{- x} x$ .

$f' (x) = - x^{2} e^{- x} + 2 x e^{- x}$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $- x^{2} e^{- x} + 2 x e^{- x}$ .

$- x^{2} e^{- x} + 2 x e^{- x}$

Step 5

Set the first derivative equal to $0$ then solve the equation $- x^{2} e^{- x} + 2 x e^{- x} = 0$ .

Tap for more steps...

Step 5.1

Set the first derivative equal to $0$ .

$- x^{2} e^{- x} + 2 x e^{- x} = 0$

Step 5.2

Factor $x e^{- x}$ out of $- x^{2} e^{- x} + 2 x e^{- x}$ .

Tap for more steps...

Step 5.2.1

Factor $x e^{- x}$ out of $- x^{2} e^{- x}$ .

$x e^{- x} (- x) + 2 x e^{- x} = 0$

Step 5.2.2

Factor $x e^{- x}$ out of $2 x e^{- x}$ .

$x e^{- x} (- x) + x e^{- x} (2) = 0$

Step 5.2.3

Factor $x e^{- x}$ out of $x e^{- x} (- x) + x e^{- x} (2)$ .

$x e^{- x} (- x + 2) = 0$

Step 5.3

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

$x = 0$

$e^{- x} = 0$

$- x + 2 = 0$

Step 5.4

Set $x$ equal to $0$ .

$x = 0$

Step 5.5

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

Tap for more steps...

Step 5.5.1

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

$e^{- x} = 0$

Step 5.5.2

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

Tap for more steps...

Step 5.5.2.1

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

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

Step 5.5.2.2

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

Undefined

Step 5.5.2.3

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

No solution

Step 5.6

Set $- x + 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 5.6.1

Set $- x + 2$ equal to $0$ .

$- x + 2 = 0$

Step 5.6.2

Solve $- x + 2 = 0$ for $x$ .

Tap for more steps...

Step 5.6.2.1

Subtract $2$ from both sides of the equation.

$- x = - 2$

Step 5.6.2.2

Divide each term in $- x = - 2$ by $- 1$ and simplify.

Tap for more steps...

Step 5.6.2.2.1

Divide each term in $- x = - 2$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 2}{- 1}$

Step 5.6.2.2.2

Simplify the left side.

Tap for more steps...

Step 5.6.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 2}{- 1}$

Step 5.6.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 2}{- 1}$

Step 5.6.2.2.3

Simplify the right side.

Tap for more steps...

Step 5.6.2.2.3.1

Divide $- 2$ by $- 1$ .

$x = 2$

Step 5.7

The final solution is all the values that make $x e^{- x} (- x + 2) = 0$ true.

$x = 0, 2$

Step 6

Find the values where the derivative is undefined.

Tap for more steps...

Step 6.1

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Step 7

Critical points to evaluate.

$x = 0, 2$

Step 8

Evaluate the second derivative at $x = 0$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

${(0)}^{2} e^{- (0)} - 4 \cdot 0 e^{- (0)} + 2 e^{- (0)}$

Step 9

Evaluate the second derivative.

Tap for more steps...

Step 9.1

Simplify each term.

Tap for more steps...

Step 9.1.1

Raising $0$ to any positive power yields $0$ .

$0 e^{- (0)} - 4 \cdot 0 e^{- (0)} + 2 e^{- (0)}$

Step 9.1.2

Multiply $- 1$ by $0$ .

$0 e^{0} - 4 \cdot 0 e^{- (0)} + 2 e^{- (0)}$

Step 9.1.3

Anything raised to $0$ is $1$ .

$0 \cdot 1 - 4 \cdot 0 e^{- (0)} + 2 e^{- (0)}$

Step 9.1.4

Multiply $0$ by $1$ .

$0 - 4 \cdot 0 e^{- (0)} + 2 e^{- (0)}$

Step 9.1.5

Multiply $- 4$ by $0$ .

$0 + 0 e^{- (0)} + 2 e^{- (0)}$

Step 9.1.6

Multiply $- 1$ by $0$ .

$0 + 0 e^{0} + 2 e^{- (0)}$

Step 9.1.7

Anything raised to $0$ is $1$ .

$0 + 0 \cdot 1 + 2 e^{- (0)}$

Step 9.1.8

Multiply $0$ by $1$ .

$0 + 0 + 2 e^{- (0)}$

Step 9.1.9

Multiply $- 1$ by $0$ .

$0 + 0 + 2 e^{0}$

Step 9.1.10

Anything raised to $0$ is $1$ .

$0 + 0 + 2 \cdot 1$

Step 9.1.11

Multiply $2$ by $1$ .

$0 + 0 + 2$

Step 9.2

Simplify by adding numbers.

Tap for more steps...

Step 9.2.1

Add $0$ and $0$ .

$0 + 2$

Step 9.2.2

Add $0$ and $2$ .

$2$

Step 10

$x = 0$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 0$ is a local minimum

Step 11

Find the y-value when $x = 0$ .

Tap for more steps...

Step 11.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = {(0)}^{2} e^{- (0)}$

Step 11.2

Simplify the result.

Tap for more steps...

Step 11.2.1

Raising $0$ to any positive power yields $0$ .

$f (0) = 0 e^{- (0)}$

Step 11.2.2

Multiply $- 1$ by $0$ .

$f (0) = 0 e^{0}$

Step 11.2.3

Anything raised to $0$ is $1$ .

$f (0) = 0 \cdot 1$

Step 11.2.4

Multiply $0$ by $1$ .

$f (0) = 0$

Step 11.2.5

The final answer is $0$ .

$y = 0$

Step 12

Evaluate the second derivative at $x = 2$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

${(2)}^{2} e^{- (2)} - 4 \cdot 2 e^{- (2)} + 2 e^{- (2)}$

Step 13

Evaluate the second derivative.

Tap for more steps...

Step 13.1

Simplify each term.

Tap for more steps...

Step 13.1.1

Raise $2$ to the power of $2$ .

$4 e^{- (2)} - 4 \cdot 2 e^{- (2)} + 2 e^{- (2)}$

Step 13.1.2

Multiply $- 1$ by $2$ .

$4 e^{- 2} - 4 \cdot 2 e^{- (2)} + 2 e^{- (2)}$

Step 13.1.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$4 \frac{1}{e^{2}} - 4 \cdot 2 e^{- (2)} + 2 e^{- (2)}$

Step 13.1.4

Combine $4$ and $\frac{1}{e^{2}}$ .

$\frac{4}{e^{2}} - 4 \cdot 2 e^{- (2)} + 2 e^{- (2)}$

Step 13.1.5

Multiply $- 4$ by $2$ .

$\frac{4}{e^{2}} - 8 e^{- (2)} + 2 e^{- (2)}$

Step 13.1.6

Multiply $- 1$ by $2$ .

$\frac{4}{e^{2}} - 8 e^{- 2} + 2 e^{- (2)}$

Step 13.1.7

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{4}{e^{2}} - 8 \frac{1}{e^{2}} + 2 e^{- (2)}$

Step 13.1.8

Combine $- 8$ and $\frac{1}{e^{2}}$ .

$\frac{4}{e^{2}} + \frac{- 8}{e^{2}} + 2 e^{- (2)}$

Step 13.1.9

Move the negative in front of the fraction.

$\frac{4}{e^{2}} - \frac{8}{e^{2}} + 2 e^{- (2)}$

Step 13.1.10

Multiply $- 1$ by $2$ .

$\frac{4}{e^{2}} - \frac{8}{e^{2}} + 2 e^{- 2}$

Step 13.1.11

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{4}{e^{2}} - \frac{8}{e^{2}} + 2 \frac{1}{e^{2}}$

Step 13.1.12

Combine $2$ and $\frac{1}{e^{2}}$ .

$\frac{4}{e^{2}} - \frac{8}{e^{2}} + \frac{2}{e^{2}}$

Step 13.2

Combine fractions.

Tap for more steps...

Step 13.2.1

Combine the numerators over the common denominator.

$\frac{4 - 8 + 2}{e^{2}}$

Step 13.2.2

Simplify the expression.

Tap for more steps...

Step 13.2.2.1

Subtract $8$ from $4$ .

$\frac{- 4 + 2}{e^{2}}$

Step 13.2.2.2

Add $- 4$ and $2$ .

$\frac{- 2}{e^{2}}$

Step 13.2.2.3

Move the negative in front of the fraction.

$- \frac{2}{e^{2}}$

Step 14

$x = 2$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = 2$ is a local maximum

Step 15

Find the y-value when $x = 2$ .

Tap for more steps...

Step 15.1

Replace the variable $x$ with $2$ in the expression.

$f (2) = {(2)}^{2} e^{- (2)}$

Step 15.2

Simplify the result.

Tap for more steps...

Step 15.2.1

Raise $2$ to the power of $2$ .

$f (2) = 4 e^{- (2)}$

Step 15.2.2

Multiply $- 1$ by $2$ .

$f (2) = 4 e^{- 2}$

Step 15.2.3

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f (2) = 4 (\frac{1}{e^{2}})$

Step 15.2.4

Combine $4$ and $\frac{1}{e^{2}}$ .

$f (2) = \frac{4}{e^{2}}$

Step 15.2.5

The final answer is $\frac{4}{e^{2}}$ .

$y = \frac{4}{e^{2}}$

Step 16

These are the local extrema for $f (x) = x^{2} e^{- x}$ .

$(0, 0)$ is a local minima

$(2, \frac{4}{e^{2}})$ is a local maxima

Step 17