Find the Concavity f(x)=e^(2x)+e^(-x)

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

Find the inflection points.

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Find the second derivative.

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Find the first derivative.

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

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

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

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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 $g (x) = 2 x$ .

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To apply the Chain Rule, set $u_{1}$ as $2 x$ .

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

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

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

Replace all occurrences of $u_{1}$ with $2 x$ .

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

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

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

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

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

Multiply $2$ by $1$ .

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

Move $2$ to the left of $e^{2 x}$ .

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

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

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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 $g (x) = - x$ .

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

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

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

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

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

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

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

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

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

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

Multiply $- 1$ by $1$ .

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

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

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

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

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

Find the second derivative.

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

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

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

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

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

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 $g (x) = 2 x$ .

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To apply the Chain Rule, set $u_{1}$ as $2 x$ .

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

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

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

Replace all occurrences of $u_{1}$ with $2 x$ .

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

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

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

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

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

Multiply $2$ by $1$ .

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

Move $2$ to the left of $e^{2 x}$ .

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

Multiply $2$ by $2$ .

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

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

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

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

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 $g (x) = - x$ .

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

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

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

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

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

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

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

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

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

$f'' (x) = 4 e^{2 x} - (e^{- x} (- 1 \cdot 1))$

Multiply $- 1$ by $1$ .

$f'' (x) = 4 e^{2 x} - (e^{- x} \cdot - 1)$

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

$f'' (x) = 4 e^{2 x} - (- 1 \cdot e^{- x})$

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

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

Multiply $- 1$ by $- 1$ .

$f'' (x) = 4 e^{2 x} + 1 e^{- x}$

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

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

The second derivative of $f (x)$ with respect to $x$ is $4 e^{2 x} + e^{- x}$ .

$4 e^{2 x} + e^{- x}$

Graph each side of the equation. The solution is the x-value of the point of intersection.

No solution

No values found that can make the second derivative equal to $0$ .

No Inflection Points

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.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

The graph is concave up because the second derivative is positive.

The graph is concave up

Find the Concavity f(x)=e^(2x)+e^(-x)

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