Calculus Examples

$e^{1 - 20 x + 5 x^{2}}$

Step 1

Write $e^{1 - 20 x + 5 x^{2}}$ as a function.

$f (x) = e^{1 - 20 x + 5 x^{2}}$

Step 2

Find the first derivative of the function.

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Step 2.1

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 $g (x) = 1 - 20 x + 5 x^{2}$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $1 - 20 x + 5 x^{2}$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [1 - 20 x + 5 x^{2}]$

Step 2.1.2

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

$e^{u} \frac{d}{d x} [1 - 20 x + 5 x^{2}]$

Step 2.1.3

Replace all occurrences of $u$ with $1 - 20 x + 5 x^{2}$ .

$e^{1 - 20 x + 5 x^{2}} \frac{d}{d x} [1 - 20 x + 5 x^{2}]$

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $1 - 20 x + 5 x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 20 x] + \frac{d}{d x} [5 x^{2}]$ .

$e^{1 - 20 x + 5 x^{2}} (\frac{d}{d x} [1] + \frac{d}{d x} [- 20 x] + \frac{d}{d x} [5 x^{2}])$

Step 2.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$e^{1 - 20 x + 5 x^{2}} (0 + \frac{d}{d x} [- 20 x] + \frac{d}{d x} [5 x^{2}])$

Step 2.2.3

Add $0$ and $\frac{d}{d x} [- 20 x]$ .

$e^{1 - 20 x + 5 x^{2}} (\frac{d}{d x} [- 20 x] + \frac{d}{d x} [5 x^{2}])$

Step 2.2.4

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 x$ with respect to $x$ is $- 20 \frac{d}{d x} [x]$ .

$e^{1 - 20 x + 5 x^{2}} (- 20 \frac{d}{d x} [x] + \frac{d}{d x} [5 x^{2}])$

Step 2.2.5

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

$e^{1 - 20 x + 5 x^{2}} (- 20 \cdot 1 + \frac{d}{d x} [5 x^{2}])$

Step 2.2.6

Multiply $- 20$ by $1$ .

$e^{1 - 20 x + 5 x^{2}} (- 20 + \frac{d}{d x} [5 x^{2}])$

Step 2.2.7

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2}$ with respect to $x$ is $5 \frac{d}{d x} [x^{2}]$ .

$e^{1 - 20 x + 5 x^{2}} (- 20 + 5 \frac{d}{d x} [x^{2}])$

Step 2.2.8

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

$e^{1 - 20 x + 5 x^{2}} (- 20 + 5 (2 x))$

Step 2.2.9

Multiply $2$ by $5$ .

$e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x)$

Step 3

Find the second derivative of the function.

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Step 3.1

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 $f (x) = e^{1 - 20 x + 5 x^{2}}$ and $g (x) = - 20 + 10 x$ .

$f'' (x) = e^{1 - 20 x + 5 x^{2}} \frac{d}{d x} (- 20 + 10 x) + (- 20 + 10 x) \frac{d}{d x} (e^{1 - 20 x + 5 x^{2}})$

Step 3.2

Differentiate.

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Step 3.2.1

By the Sum Rule, the derivative of $- 20 + 10 x$ with respect to $x$ is $\frac{d}{d x} [- 20] + \frac{d}{d x} [10 x]$ .

$f'' (x) = e^{1 - 20 x + 5 x^{2}} (\frac{d}{d x} (- 20) + \frac{d}{d x} (10 x)) + (- 20 + 10 x) \frac{d}{d x} (e^{1 - 20 x + 5 x^{2}})$

Step 3.2.2

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20$ with respect to $x$ is $0$ .

$f'' (x) = e^{1 - 20 x + 5 x^{2}} (0 + \frac{d}{d x} (10 x)) + (- 20 + 10 x) \frac{d}{d x} (e^{1 - 20 x + 5 x^{2}})$

Step 3.2.3

Add $0$ and $\frac{d}{d x} [10 x]$ .

$f'' (x) = e^{1 - 20 x + 5 x^{2}} \frac{d}{d x} (10 x) + (- 20 + 10 x) \frac{d}{d x} (e^{1 - 20 x + 5 x^{2}})$

Step 3.2.4

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

$f'' (x) = e^{1 - 20 x + 5 x^{2}} (10 \frac{d}{d x} (x)) + (- 20 + 10 x) \frac{d}{d x} (e^{1 - 20 x + 5 x^{2}})$

Step 3.2.5

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

$f'' (x) = e^{1 - 20 x + 5 x^{2}} (10 \cdot 1) + (- 20 + 10 x) \frac{d}{d x} (e^{1 - 20 x + 5 x^{2}})$

Step 3.2.6

Simplify the expression.

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Step 3.2.6.1

Multiply $10$ by $1$ .

$f'' (x) = e^{1 - 20 x + 5 x^{2}} \cdot 10 + (- 20 + 10 x) \frac{d}{d x} (e^{1 - 20 x + 5 x^{2}})$

Step 3.2.6.2

Move $10$ to the left of $e^{1 - 20 x + 5 x^{2}}$ .

$f'' (x) = 10 \cdot e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) \frac{d}{d x} (e^{1 - 20 x + 5 x^{2}})$

Step 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 $g (x) = 1 - 20 x + 5 x^{2}$ .

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Step 3.3.1

To apply the Chain Rule, set $u$ as $1 - 20 x + 5 x^{2}$ .

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) (\frac{d}{d u} (e^{u}) \frac{d}{d x} (1 - 20 x + 5 x^{2}))$

Step 3.3.2

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

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) (e^{u} \frac{d}{d x} (1 - 20 x + 5 x^{2}))$

Step 3.3.3

Replace all occurrences of $u$ with $1 - 20 x + 5 x^{2}$ .

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) (e^{1 - 20 x + 5 x^{2}} \frac{d}{d x} (1 - 20 x + 5 x^{2}))$

Step 3.4

Differentiate.

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Step 3.4.1

By the Sum Rule, the derivative of $1 - 20 x + 5 x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 20 x] + \frac{d}{d x} [5 x^{2}]$ .

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) e^{1 - 20 x + 5 x^{2}} (\frac{d}{d x} (1) + \frac{d}{d x} (- 20 x) + \frac{d}{d x} (5 x^{2}))$

Step 3.4.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) e^{1 - 20 x + 5 x^{2}} (0 + \frac{d}{d x} (- 20 x) + \frac{d}{d x} (5 x^{2}))$

Step 3.4.3

Add $0$ and $\frac{d}{d x} [- 20 x]$ .

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) e^{1 - 20 x + 5 x^{2}} (\frac{d}{d x} (- 20 x) + \frac{d}{d x} (5 x^{2}))$

Step 3.4.4

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 x$ with respect to $x$ is $- 20 \frac{d}{d x} [x]$ .

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) e^{1 - 20 x + 5 x^{2}} (- 20 \frac{d}{d x} x + \frac{d}{d x} (5 x^{2}))$

Step 3.4.5

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

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) e^{1 - 20 x + 5 x^{2}} (- 20 \cdot 1 + \frac{d}{d x} (5 x^{2}))$

Step 3.4.6

Multiply $- 20$ by $1$ .

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) e^{1 - 20 x + 5 x^{2}} (- 20 + \frac{d}{d x} (5 x^{2}))$

Step 3.4.7

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2}$ with respect to $x$ is $5 \frac{d}{d x} [x^{2}]$ .

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) e^{1 - 20 x + 5 x^{2}} (- 20 + 5 \frac{d}{d x} (x^{2}))$

Step 3.4.8

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

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) e^{1 - 20 x + 5 x^{2}} (- 20 + 5 (2 x))$

Step 3.4.9

Multiply $2$ by $5$ .

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x)$

Step 3.5

Raise $- 20 + 10 x$ to the power of $1$ .

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) (- 20 + 10 x) e^{1 - 20 x + 5 x^{2}}$

Step 3.6

Raise $- 20 + 10 x$ to the power of $1$ .

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + (- 20 + 10 x) (- 20 + 10 x) e^{1 - 20 x + 5 x^{2}}$

Step 3.7

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + {(- 20 + 10 x)}^{1 + 1} e^{1 - 20 x + 5 x^{2}}$

Step 3.8

Add $1$ and $1$ .

$f'' (x) = 10 e^{1 - 20 x + 5 x^{2}} + {(- 20 + 10 x)}^{2} e^{1 - 20 x + 5 x^{2}}$

Step 3.9

Reorder terms.

$f'' (x) = e^{1 - 20 x + 5 x^{2}} {(- 20 + 10 x)}^{2} + 10 e^{1 - 20 x + 5 x^{2}}$

Step 4

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

$e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x) = 0$

Step 5

Find the first derivative.

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Step 5.1

Find the first derivative.

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Step 5.1.1

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 $g (x) = 1 - 20 x + 5 x^{2}$ .

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Step 5.1.1.1

To apply the Chain Rule, set $u$ as $1 - 20 x + 5 x^{2}$ .

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [1 - 20 x + 5 x^{2}]$

Step 5.1.1.2

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

$e^{u} \frac{d}{d x} [1 - 20 x + 5 x^{2}]$

Step 5.1.1.3

Replace all occurrences of $u$ with $1 - 20 x + 5 x^{2}$ .

$e^{1 - 20 x + 5 x^{2}} \frac{d}{d x} [1 - 20 x + 5 x^{2}]$

Step 5.1.2

Differentiate.

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Step 5.1.2.1

By the Sum Rule, the derivative of $1 - 20 x + 5 x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- 20 x] + \frac{d}{d x} [5 x^{2}]$ .

$e^{1 - 20 x + 5 x^{2}} (\frac{d}{d x} [1] + \frac{d}{d x} [- 20 x] + \frac{d}{d x} [5 x^{2}])$

Step 5.1.2.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$e^{1 - 20 x + 5 x^{2}} (0 + \frac{d}{d x} [- 20 x] + \frac{d}{d x} [5 x^{2}])$

Step 5.1.2.3

Add $0$ and $\frac{d}{d x} [- 20 x]$ .

$e^{1 - 20 x + 5 x^{2}} (\frac{d}{d x} [- 20 x] + \frac{d}{d x} [5 x^{2}])$

Step 5.1.2.4

Since $- 20$ is constant with respect to $x$ , the derivative of $- 20 x$ with respect to $x$ is $- 20 \frac{d}{d x} [x]$ .

$e^{1 - 20 x + 5 x^{2}} (- 20 \frac{d}{d x} [x] + \frac{d}{d x} [5 x^{2}])$

Step 5.1.2.5

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

$e^{1 - 20 x + 5 x^{2}} (- 20 \cdot 1 + \frac{d}{d x} [5 x^{2}])$

Step 5.1.2.6

Multiply $- 20$ by $1$ .

$e^{1 - 20 x + 5 x^{2}} (- 20 + \frac{d}{d x} [5 x^{2}])$

Step 5.1.2.7

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2}$ with respect to $x$ is $5 \frac{d}{d x} [x^{2}]$ .

$e^{1 - 20 x + 5 x^{2}} (- 20 + 5 \frac{d}{d x} [x^{2}])$

Step 5.1.2.8

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

$e^{1 - 20 x + 5 x^{2}} (- 20 + 5 (2 x))$

Step 5.1.2.9

Multiply $2$ by $5$ .

$f' (x) = e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x)$

Step 5.2

The first derivative of $f (x)$ with respect to $x$ is $e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x)$ .

$e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x)$

Step 6

Set the first derivative equal to $0$ then solve the equation $e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x) = 0$ .

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Step 6.1

Set the first derivative equal to $0$ .

$e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x) = 0$

Step 6.2

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

$e^{1 - 20 x + 5 x^{2}} = 0$

$- 20 + 10 x = 0$

Step 6.3

Set $e^{1 - 20 x + 5 x^{2}}$ equal to $0$ and solve for $x$ .

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Step 6.3.1

Set $e^{1 - 20 x + 5 x^{2}}$ equal to $0$ .

$e^{1 - 20 x + 5 x^{2}} = 0$

Step 6.3.2

Solve $e^{1 - 20 x + 5 x^{2}} = 0$ for $x$ .

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Step 6.3.2.1

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

$\ln (e^{1 - 20 x + 5 x^{2}}) = \ln (0)$

Step 6.3.2.2

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

Undefined

Step 6.3.2.3

There is no solution for $e^{1 - 20 x + 5 x^{2}} = 0$

No solution

Step 6.4

Set $- 20 + 10 x$ equal to $0$ and solve for $x$ .

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Step 6.4.1

Set $- 20 + 10 x$ equal to $0$ .

$- 20 + 10 x = 0$

Step 6.4.2

Solve $- 20 + 10 x = 0$ for $x$ .

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Step 6.4.2.1

Add $20$ to both sides of the equation.

$10 x = 20$

Step 6.4.2.2

Divide each term in $10 x = 20$ by $10$ and simplify.

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Step 6.4.2.2.1

Divide each term in $10 x = 20$ by $10$ .

$\frac{10 x}{10} = \frac{20}{10}$

Step 6.4.2.2.2

Simplify the left side.

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Step 6.4.2.2.2.1

Cancel the common factor of $10$ .

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Step 6.4.2.2.2.1.1

Cancel the common factor.

$\frac{10 x}{10} = \frac{20}{10}$

Step 6.4.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{20}{10}$

Step 6.4.2.2.3

Simplify the right side.

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Step 6.4.2.2.3.1

Divide $20$ by $10$ .

$x = 2$

Step 6.5

The final solution is all the values that make $e^{1 - 20 x + 5 x^{2}} (- 20 + 10 x) = 0$ true.

$x = 2$

Step 7

Find the values where the derivative is undefined.

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Step 7.1

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 8

Critical points to evaluate.

$x = 2$

Step 9

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.

$e^{1 - 20 \cdot 2 + 5 {(2)}^{2}} {(- 20 + 10 (2))}^{2} + 10 e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 10

Evaluate the second derivative.

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Step 10.1

Simplify each term.

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Step 10.1.1

Simplify each term.

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Step 10.1.1.1

Multiply $- 20$ by $2$ .

$e^{1 - 40 + 5 {(2)}^{2}} {(- 20 + 10 (2))}^{2} + 10 e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 10.1.1.2

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

$e^{1 - 40 + 5 \cdot 4} {(- 20 + 10 (2))}^{2} + 10 e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 10.1.1.3

Multiply $5$ by $4$ .

$e^{1 - 40 + 20} {(- 20 + 10 (2))}^{2} + 10 e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 10.1.2

Subtract $40$ from $1$ .

$e^{- 39 + 20} {(- 20 + 10 (2))}^{2} + 10 e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 10.1.3

Add $- 39$ and $20$ .

$e^{- 19} {(- 20 + 10 (2))}^{2} + 10 e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 10.1.4

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

$\frac{1}{e^{19}} {(- 20 + 10 (2))}^{2} + 10 e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 10.1.5

Multiply $10$ by $2$ .

$\frac{1}{e^{19}} {(- 20 + 20)}^{2} + 10 e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 10.1.6

Add $- 20$ and $20$ .

$\frac{1}{e^{19}} \cdot 0^{2} + 10 e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 10.1.7

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

$\frac{1}{e^{19}} \cdot 0 + 10 e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 10.1.8

Multiply $\frac{1}{e^{19}}$ by $0$ .

$0 + 10 e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 10.1.9

Simplify each term.

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Step 10.1.9.1

Multiply $- 20$ by $2$ .

$0 + 10 e^{1 - 40 + 5 {(2)}^{2}}$

Step 10.1.9.2

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

$0 + 10 e^{1 - 40 + 5 \cdot 4}$

Step 10.1.9.3

Multiply $5$ by $4$ .

$0 + 10 e^{1 - 40 + 20}$

Step 10.1.10

Subtract $40$ from $1$ .

$0 + 10 e^{- 39 + 20}$

Step 10.1.11

Add $- 39$ and $20$ .

$0 + 10 e^{- 19}$

Step 10.1.12

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

$0 + 10 \frac{1}{e^{19}}$

Step 10.1.13

Combine $10$ and $\frac{1}{e^{19}}$ .

$0 + \frac{10}{e^{19}}$

Step 10.2

Add $0$ and $\frac{10}{e^{19}}$ .

$\frac{10}{e^{19}}$

Step 11

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

$x = 2$ is a local minimum

Step 12

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

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Step 12.1

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

$f (2) = e^{1 - 20 \cdot 2 + 5 {(2)}^{2}}$

Step 12.2

Simplify the result.

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Step 12.2.1

Simplify each term.

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Step 12.2.1.1

Multiply $- 20$ by $2$ .

$f (2) = e^{1 - 40 + 5 {(2)}^{2}}$

Step 12.2.1.2

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

$f (2) = e^{1 - 40 + 5 \cdot 4}$

Step 12.2.1.3

Multiply $5$ by $4$ .

$f (2) = e^{1 - 40 + 20}$

Step 12.2.2

Simplify by adding and subtracting.

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Step 12.2.2.1

Subtract $40$ from $1$ .

$f (2) = e^{- 39 + 20}$

Step 12.2.2.2

Add $- 39$ and $20$ .

$f (2) = e^{- 19}$

Step 12.2.3

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

$f (2) = \frac{1}{e^{19}}$

Step 12.2.4

The final answer is $\frac{1}{e^{19}}$ .

$y = \frac{1}{e^{19}}$

Step 13

These are the local extrema for $f (x) = e^{1 - 20 x + 5 x^{2}}$ .

$(2, \frac{1}{e^{19}})$ is a local minima

Step 14