Find the Inflection Points 10xe^(-4x)

$10 x e^{- 4 x}$

Step 1

Write $10 x e^{- 4 x}$ as a function.

$f (x) = 10 x e^{- 4 x}$

Step 2

Find the second derivative.

Find the first derivative.

Since $10$ is constant with respect to $x$ , the derivative of $10 x e^{- 4 x}$ with respect to $x$ is $10 \frac{d}{d x} [x e^{- 4 x}]$ .

$f' (x) = 10 \frac{d}{d x} (x e^{- 4 x})$

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 $g (x) = e^{- 4 x}$ .

$f' (x) = 10 (x \frac{d}{d x} (e^{- 4 x}) + e^{- 4 x} \frac{d}{d x} (x))$

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 $g (x) = - 4 x$ .

To apply the Chain Rule, set $u$ as $- 4 x$ .

$f' (x) = 10 (x (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- 4 x)) + e^{- 4 x} \frac{d}{d x} (x))$

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

$f' (x) = 10 (x (e^{u} \frac{d}{d x} (- 4 x)) + e^{- 4 x} \frac{d}{d x} (x))$

Replace all occurrences of $u$ with $- 4 x$ .

$f' (x) = 10 (x (e^{- 4 x} \frac{d}{d x} (- 4 x)) + e^{- 4 x} \frac{d}{d x} (x))$

Differentiate.

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$f' (x) = 10 (x (e^{- 4 x} (- 4 \frac{d}{d x} x)) + e^{- 4 x} \frac{d}{d x} (x))$

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

$f' (x) = 10 (x (e^{- 4 x} (- 4 \cdot 1)) + e^{- 4 x} \frac{d}{d x} (x))$

Simplify the expression.

Multiply $- 4$ by $1$ .

$f' (x) = 10 (x (e^{- 4 x} \cdot - 4) + e^{- 4 x} \frac{d}{d x} (x))$

Move $- 4$ to the left of $e^{- 4 x}$ .

$f' (x) = 10 (x (- 4 \cdot e^{- 4 x}) + e^{- 4 x} \frac{d}{d x} (x))$

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

$f' (x) = 10 (x (- 4 e^{- 4 x}) + e^{- 4 x} \cdot 1)$

Multiply $e^{- 4 x}$ by $1$ .

$f' (x) = 10 (x (- 4 e^{- 4 x}) + e^{- 4 x})$

Simplify.

Apply the distributive property.

$f' (x) = 10 (x (- 4 e^{- 4 x})) + 10 e^{- 4 x}$

Multiply $- 4$ by $10$ .

$f' (x) = - 40 x e^{- 4 x} + 10 e^{- 4 x}$

Reorder terms.

$f' (x) = - 40 e^{- 4 x} x + 10 e^{- 4 x}$

Reorder factors in $- 40 e^{- 4 x} x + 10 e^{- 4 x}$ .

$f' (x) = - 40 x e^{- 4 x} + 10 e^{- 4 x}$

Find the second derivative.

By the Sum Rule, the derivative of $- 40 x e^{- 4 x} + 10 e^{- 4 x}$ with respect to $x$ is $\frac{d}{d x} [- 40 x e^{- 4 x}] + \frac{d}{d x} [10 e^{- 4 x}]$ .

$f'' (x) = \frac{d}{d x} (- 40 x e^{- 4 x}) + \frac{d}{d x} (10 e^{- 4 x})$

Evaluate $\frac{d}{d x} [- 40 x e^{- 4 x}]$ .

Since $- 40$ is constant with respect to $x$ , the derivative of $- 40 x e^{- 4 x}$ with respect to $x$ is $- 40 \frac{d}{d x} [x e^{- 4 x}]$ .

$f'' (x) = - 40 \frac{d}{d x} (x e^{- 4 x}) + \frac{d}{d x} (10 e^{- 4 x})$

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 $g (x) = e^{- 4 x}$ .

$f'' (x) = - 40 (x \frac{d}{d x} (e^{- 4 x}) + e^{- 4 x} \frac{d}{d x} (x)) + \frac{d}{d x} (10 e^{- 4 x})$

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 $g (x) = - 4 x$ .

To apply the Chain Rule, set $u_{1}$ as $- 4 x$ .

$f'' (x) = - 40 (x (\frac{d}{d u (1)} (e^{u_{1}}) \frac{d}{d x} (- 4 x)) + e^{- 4 x} \frac{d}{d x} (x)) + \frac{d}{d x} (10 e^{- 4 x})$

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

$f'' (x) = - 40 (x (e^{u_{1}} \frac{d}{d x} (- 4 x)) + e^{- 4 x} \frac{d}{d x} (x)) + \frac{d}{d x} (10 e^{- 4 x})$

Replace all occurrences of $u_{1}$ with $- 4 x$ .

$f'' (x) = - 40 (x (e^{- 4 x} \frac{d}{d x} (- 4 x)) + e^{- 4 x} \frac{d}{d x} (x)) + \frac{d}{d x} (10 e^{- 4 x})$

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$f'' (x) = - 40 (x (e^{- 4 x} (- 4 \frac{d}{d x} x)) + e^{- 4 x} \frac{d}{d x} (x)) + \frac{d}{d x} (10 e^{- 4 x})$

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

$f'' (x) = - 40 (x (e^{- 4 x} (- 4 \cdot 1)) + e^{- 4 x} \frac{d}{d x} (x)) + \frac{d}{d x} (10 e^{- 4 x})$

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

$f'' (x) = - 40 (x (e^{- 4 x} (- 4 \cdot 1)) + e^{- 4 x} \cdot 1) + \frac{d}{d x} (10 e^{- 4 x})$

Multiply $- 4$ by $1$ .

$f'' (x) = - 40 (x (e^{- 4 x} \cdot - 4) + e^{- 4 x} \cdot 1) + \frac{d}{d x} (10 e^{- 4 x})$

Move $- 4$ to the left of $e^{- 4 x}$ .

$f'' (x) = - 40 (x (- 4 \cdot e^{- 4 x}) + e^{- 4 x} \cdot 1) + \frac{d}{d x} (10 e^{- 4 x})$

Multiply $e^{- 4 x}$ by $1$ .

$f'' (x) = - 40 (x (- 4 e^{- 4 x}) + e^{- 4 x}) + \frac{d}{d x} (10 e^{- 4 x})$

Evaluate $\frac{d}{d x} [10 e^{- 4 x}]$ .

Since $10$ is constant with respect to $x$ , the derivative of $10 e^{- 4 x}$ with respect to $x$ is $10 \frac{d}{d x} [e^{- 4 x}]$ .

$f'' (x) = - 40 (x (- 4 e^{- 4 x}) + e^{- 4 x}) + 10 \frac{d}{d x} (e^{- 4 x})$

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 $g (x) = - 4 x$ .

To apply the Chain Rule, set $u_{2}$ as $- 4 x$ .

$f'' (x) = - 40 (x (- 4 e^{- 4 x}) + e^{- 4 x}) + 10 (\frac{d}{d u (2)} (e^{u_{2}}) \frac{d}{d x} (- 4 x))$

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

$f'' (x) = - 40 (x (- 4 e^{- 4 x}) + e^{- 4 x}) + 10 (e^{u_{2}} \frac{d}{d x} (- 4 x))$

Replace all occurrences of $u_{2}$ with $- 4 x$ .

$f'' (x) = - 40 (x (- 4 e^{- 4 x}) + e^{- 4 x}) + 10 (e^{- 4 x} \frac{d}{d x} (- 4 x))$

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 x$ with respect to $x$ is $- 4 \frac{d}{d x} [x]$ .

$f'' (x) = - 40 (x (- 4 e^{- 4 x}) + e^{- 4 x}) + 10 (e^{- 4 x} (- 4 \frac{d}{d x} x))$

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

$f'' (x) = - 40 (x (- 4 e^{- 4 x}) + e^{- 4 x}) + 10 (e^{- 4 x} (- 4 \cdot 1))$

Multiply $- 4$ by $1$ .

$f'' (x) = - 40 (x (- 4 e^{- 4 x}) + e^{- 4 x}) + 10 (e^{- 4 x} \cdot - 4)$

Move $- 4$ to the left of $e^{- 4 x}$ .

$f'' (x) = - 40 (x (- 4 e^{- 4 x}) + e^{- 4 x}) + 10 (- 4 \cdot e^{- 4 x})$

Multiply $- 4$ by $10$ .

$f'' (x) = - 40 (x (- 4 e^{- 4 x}) + e^{- 4 x}) - 40 e^{- 4 x}$

Simplify.

Apply the distributive property.

$f'' (x) = - 40 (x (- 4 e^{- 4 x})) - 40 e^{- 4 x} - 40 e^{- 4 x}$

Combine terms.

Multiply $- 4$ by $- 40$ .

$f'' (x) = 160 (x (e^{- 4 x})) - 40 e^{- 4 x} - 40 e^{- 4 x}$

Subtract $40 e^{- 4 x}$ from $- 40 e^{- 4 x}$ .

$f'' (x) = 160 x e^{- 4 x} - 80 e^{- 4 x}$

Reorder terms.

$f'' (x) = 160 e^{- 4 x} x - 80 e^{- 4 x}$

Reorder factors in $160 e^{- 4 x} x - 80 e^{- 4 x}$ .

$f'' (x) = 160 x e^{- 4 x} - 80 e^{- 4 x}$

The second derivative of $f (x)$ with respect to $x$ is $160 x e^{- 4 x} - 80 e^{- 4 x}$ .

$160 x e^{- 4 x} - 80 e^{- 4 x}$

Step 3

Set the second derivative equal to $0$ then solve the equation $160 x e^{- 4 x} - 80 e^{- 4 x} = 0$ .

Set the second derivative equal to $0$ .

$160 x e^{- 4 x} - 80 e^{- 4 x} = 0$

Factor $80 e^{- 4 x}$ out of $160 x e^{- 4 x} - 80 e^{- 4 x}$ .

Factor $80 e^{- 4 x}$ out of $160 x e^{- 4 x}$ .

$80 e^{- 4 x} (2 x) - 80 e^{- 4 x} = 0$

Factor $80 e^{- 4 x}$ out of $- 80 e^{- 4 x}$ .

$80 e^{- 4 x} (2 x) + 80 e^{- 4 x} (- 1) = 0$

Factor $80 e^{- 4 x}$ out of $80 e^{- 4 x} (2 x) + 80 e^{- 4 x} (- 1)$ .

$80 e^{- 4 x} (2 x - 1) = 0$

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

$e^{- 4 x} = 0$

$2 x - 1 = 0$

Set $e^{- 4 x}$ equal to $0$ and solve for $x$ .

Set $e^{- 4 x}$ equal to $0$ .

$e^{- 4 x} = 0$

Solve $e^{- 4 x} = 0$ for $x$ .

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

$\ln (e^{- 4 x}) = \ln (0)$

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

Undefined

There is no solution for $e^{- 4 x} = 0$

No solution

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

The final solution is all the values that make $80 e^{- 4 x} (2 x - 1) = 0$ true.

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

Step 4

Find the points where the second derivative is $0$ .

Substitute $\frac{1}{2}$ in $f (x) = 10 x e^{- 4 x}$ to find the value of $y$ .

Replace the variable $x$ with $\frac{1}{2}$ in the expression.

$f (\frac{1}{2}) = 10 (\frac{1}{2}) \cdot e^{- 4 (\frac{1}{2})}$

Simplify the result.

Cancel the common factor of $2$ .

Factor $2$ out of $10$ .

$f (\frac{1}{2}) = 2 (5) (\frac{1}{2}) \cdot e^{- 4 (\frac{1}{2})}$

Cancel the common factor.

$f (\frac{1}{2}) = 2 \cdot (5 (\frac{1}{2})) \cdot e^{- 4 (\frac{1}{2})}$

Rewrite the expression.

$f (\frac{1}{2}) = 5 \cdot e^{- 4 (\frac{1}{2})}$

Cancel the common factor of $2$ .

Factor $2$ out of $- 4$ .

$f (\frac{1}{2}) = 5 \cdot e^{2 (- 2) (\frac{1}{2})}$

Cancel the common factor.

$f (\frac{1}{2}) = 5 \cdot e^{2 \cdot (- 2 (\frac{1}{2}))}$

Rewrite the expression.

$f (\frac{1}{2}) = 5 \cdot e^{- 2}$

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

$f (\frac{1}{2}) = 5 \cdot \frac{1}{e^{2}}$

Combine $5$ and $\frac{1}{e^{2}}$ .

$f (\frac{1}{2}) = \frac{5}{e^{2}}$

The final answer is $\frac{5}{e^{2}}$ .

$\frac{5}{e^{2}}$

The point found by substituting $\frac{1}{2}$ in $f (x) = 10 x e^{- 4 x}$ is $(\frac{1}{2}, \frac{5}{e^{2}})$ . This point can be an inflection point.

$(\frac{1}{2}, \frac{5}{e^{2}})$

Step 5

Split $(- \infty, \infty)$ into intervals around the points that could potentially be inflection points.

$(- \infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)$

Step 6

Substitute a value from the interval $(- \infty, \frac{1}{2})$ into the second derivative to determine if it is increasing or decreasing.

Replace the variable $x$ with $0.4$ in the expression.

$f'' (0.4) = 160 (0.4) \cdot e^{- 4 \cdot 0.4} - 80 e^{- 4 \cdot 0.4}$

Simplify the result.

Simplify each term.

Multiply $160$ by $0.4$ .

$f'' (0.4) = 64 \cdot e^{- 4 \cdot 0.4} - 80 e^{- 4 \cdot 0.4}$

Multiply $- 4$ by $0.4$ .

$f'' (0.4) = 64 \cdot e^{- 1.6} - 80 e^{- 4 \cdot 0.4}$

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

$f'' (0.4) = 64 \cdot \frac{1}{e^{1.6}} - 80 e^{- 4 \cdot 0.4}$

Combine $64$ and $\frac{1}{e^{1.6}}$ .

$f'' (0.4) = \frac{64}{e^{1.6}} - 80 e^{- 4 \cdot 0.4}$

Replace $e$ with an approximation.

$f'' (0.4) = \frac{64}{{2.71828182}^{1.6}} - 80 e^{- 4 \cdot 0.4}$

Raise $2.71828182$ to the power of $1.6$ .

$f'' (0.4) = \frac{64}{4.95303242} - 80 e^{- 4 \cdot 0.4}$

Divide $64$ by $4.95303242$ .

$f'' (0.4) = 12.92137715 - 80 e^{- 4 \cdot 0.4}$

Multiply $- 4$ by $0.4$ .

$f'' (0.4) = 12.92137715 - 80 e^{- 1.6}$

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

$f'' (0.4) = 12.92137715 - 80 \frac{1}{e^{1.6}}$

Combine $- 80$ and $\frac{1}{e^{1.6}}$ .

$f'' (0.4) = 12.92137715 + \frac{- 80}{e^{1.6}}$

Move the negative in front of the fraction.

$f'' (0.4) = 12.92137715 - \frac{80}{e^{1.6}}$

Subtract $\frac{80}{e^{1.6}}$ from $12.92137715$ .

$f'' (0.4) = - 3.23034428$

The final answer is $- 3.23034428$ .

$- 3.23034428$

At $0.4$ , the second derivative is $- 3.23034428$ . Since this is negative, the second derivative is decreasing on the interval $(- \infty, \frac{1}{2})$

Decreasing on $(- \infty, \frac{1}{2})$ since $f'' (x) < 0$

Step 7

Substitute a value from the interval $(\frac{1}{2}, \infty)$ into the second derivative to determine if it is increasing or decreasing.

Replace the variable $x$ with $0.6$ in the expression.

$f'' (0.6) = 160 (0.6) \cdot e^{- 4 \cdot 0.6} - 80 e^{- 4 \cdot 0.6}$

Simplify the result.

Simplify each term.

Multiply $160$ by $0.6$ .

$f'' (0.6) = 96 \cdot e^{- 4 \cdot 0.6} - 80 e^{- 4 \cdot 0.6}$

Multiply $- 4$ by $0.6$ .

$f'' (0.6) = 96 \cdot e^{- 2.4} - 80 e^{- 4 \cdot 0.6}$

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

$f'' (0.6) = 96 \cdot \frac{1}{e^{2.4}} - 80 e^{- 4 \cdot 0.6}$

Combine $96$ and $\frac{1}{e^{2.4}}$ .

$f'' (0.6) = \frac{96}{e^{2.4}} - 80 e^{- 4 \cdot 0.6}$

Replace $e$ with an approximation.

$f'' (0.6) = \frac{96}{{2.71828182}^{2.4}} - 80 e^{- 4 \cdot 0.6}$

Raise $2.71828182$ to the power of $2.4$ .

$f'' (0.6) = \frac{96}{11.02317638} - 80 e^{- 4 \cdot 0.6}$

Divide $96$ by $11.02317638$ .

$f'' (0.6) = 8.70892351 - 80 e^{- 4 \cdot 0.6}$

Multiply $- 4$ by $0.6$ .

$f'' (0.6) = 8.70892351 - 80 e^{- 2.4}$

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

$f'' (0.6) = 8.70892351 - 80 \frac{1}{e^{2.4}}$

Combine $- 80$ and $\frac{1}{e^{2.4}}$ .

$f'' (0.6) = 8.70892351 + \frac{- 80}{e^{2.4}}$

Move the negative in front of the fraction.

$f'' (0.6) = 8.70892351 - \frac{80}{e^{2.4}}$

Subtract $\frac{80}{e^{2.4}}$ from $8.70892351$ .

$f'' (0.6) = 1.45148725$

The final answer is $1.45148725$ .

$1.45148725$

At $0.6$ , the second derivative is $1.45148725$ . Since this is positive, the second derivative is increasing on the interval $(\frac{1}{2}, \infty)$ .

Increasing on $(\frac{1}{2}, \infty)$ since $f'' (x) > 0$

Step 8

An inflection point is a point on a curve at which the concavity changes sign from plus to minus or from minus to plus. The inflection point in this case is $(\frac{1}{2}, \frac{5}{e^{2}})$ .

$(\frac{1}{2}, \frac{5}{e^{2}})$

Step 9

Find the Inflection Points 10xe^(-4x)

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