Calculus Examples

$f (x) = 2 x^{2} - 16 x + 27$

Step 1

Find the first derivative of the function.

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Step 1.1

By the Sum Rule, the derivative of $2 x^{2} - 16 x + 27$ with respect to $x$ is $\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 16 x] + \frac{d}{d x} [27]$ .

$\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 16 x] + \frac{d}{d x} [27]$

Step 1.2

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

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Step 1.2.1

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

$2 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16 x] + \frac{d}{d x} [27]$

Step 1.2.2

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

$2 (2 x) + \frac{d}{d x} [- 16 x] + \frac{d}{d x} [27]$

Step 1.2.3

Multiply $2$ by $2$ .

$4 x + \frac{d}{d x} [- 16 x] + \frac{d}{d x} [27]$

Step 1.3

Evaluate $\frac{d}{d x} [- 16 x]$ .

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Step 1.3.1

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16 x$ with respect to $x$ is $- 16 \frac{d}{d x} [x]$ .

$4 x - 16 \frac{d}{d x} [x] + \frac{d}{d x} [27]$

Step 1.3.2

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

$4 x - 16 \cdot 1 + \frac{d}{d x} [27]$

Step 1.3.3

Multiply $- 16$ by $1$ .

$4 x - 16 + \frac{d}{d x} [27]$

Step 1.4

Differentiate using the Constant Rule.

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Step 1.4.1

Since $27$ is constant with respect to $x$ , the derivative of $27$ with respect to $x$ is $0$ .

$4 x - 16 + 0$

Step 1.4.2

Add $4 x - 16$ and $0$ .

$4 x - 16$

Step 2

Find the second derivative of the function.

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Step 2.1

By the Sum Rule, the derivative of $4 x - 16$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [- 16]$ .

$f'' (x) = \frac{d}{d x} (4 x) + \frac{d}{d x} (- 16)$

Step 2.2

Evaluate $\frac{d}{d x} [4 x]$ .

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Step 2.2.1

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

$f'' (x) = 4 \frac{d}{d x} (x) + \frac{d}{d x} (- 16)$

Step 2.2.2

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

$f'' (x) = 4 \cdot 1 + \frac{d}{d x} (- 16)$

Step 2.2.3

Multiply $4$ by $1$ .

$f'' (x) = 4 + \frac{d}{d x} (- 16)$

Step 2.3

Differentiate using the Constant Rule.

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Step 2.3.1

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16$ with respect to $x$ is $0$ .

$f'' (x) = 4 + 0$

Step 2.3.2

Add $4$ and $0$ .

$f'' (x) = 4$

Step 3

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

$4 x - 16 = 0$

Step 4

Find the first derivative.

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Step 4.1

Find the first derivative.

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Step 4.1.1

By the Sum Rule, the derivative of $2 x^{2} - 16 x + 27$ with respect to $x$ is $\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 16 x] + \frac{d}{d x} [27]$ .

$\frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [- 16 x] + \frac{d}{d x} [27]$

Step 4.1.2

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

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Step 4.1.2.1

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

$2 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 16 x] + \frac{d}{d x} [27]$

Step 4.1.2.2

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

$2 (2 x) + \frac{d}{d x} [- 16 x] + \frac{d}{d x} [27]$

Step 4.1.2.3

Multiply $2$ by $2$ .

$4 x + \frac{d}{d x} [- 16 x] + \frac{d}{d x} [27]$

Step 4.1.3

Evaluate $\frac{d}{d x} [- 16 x]$ .

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Step 4.1.3.1

Since $- 16$ is constant with respect to $x$ , the derivative of $- 16 x$ with respect to $x$ is $- 16 \frac{d}{d x} [x]$ .

$4 x - 16 \frac{d}{d x} [x] + \frac{d}{d x} [27]$

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 $n = 1$ .

$4 x - 16 \cdot 1 + \frac{d}{d x} [27]$

Step 4.1.3.3

Multiply $- 16$ by $1$ .

$4 x - 16 + \frac{d}{d x} [27]$

Step 4.1.4

Differentiate using the Constant Rule.

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Step 4.1.4.1

Since $27$ is constant with respect to $x$ , the derivative of $27$ with respect to $x$ is $0$ .

$4 x - 16 + 0$

Step 4.1.4.2

Add $4 x - 16$ and $0$ .

$f' (x) = 4 x - 16$

Step 4.2

The first derivative of $f (x)$ with respect to $x$ is $4 x - 16$ .

$4 x - 16$

Step 5

Set the first derivative equal to $0$ then solve the equation $4 x - 16 = 0$ .

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Step 5.1

Set the first derivative equal to $0$ .

$4 x - 16 = 0$

Step 5.2

Add $16$ to both sides of the equation.

$4 x = 16$

Step 5.3

Divide each term in $4 x = 16$ by $4$ and simplify.

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Step 5.3.1

Divide each term in $4 x = 16$ by $4$ .

$\frac{4 x}{4} = \frac{16}{4}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $4$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{16}{4}$

Step 5.3.2.1.2

Divide $x$ by $1$ .

$x = \frac{16}{4}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Divide $16$ by $4$ .

$x = 4$

Step 6

Find the values where the derivative is undefined.

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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.

$x = 4$

Step 8

Evaluate the second derivative at $x = 4$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$4$

Step 9

$x = 4$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 4$ is a local minimum

Step 10

Find the y-value when $x = 4$ .

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Step 10.1

Replace the variable $x$ with $4$ in the expression.

$f (4) = 2 {(4)}^{2} - 16 \cdot 4 + 27$

Step 10.2

Simplify the result.

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Step 10.2.1

Simplify each term.

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Step 10.2.1.1

Raise $4$ to the power of $2$ .

$f (4) = 2 \cdot 16 - 16 \cdot 4 + 27$

Step 10.2.1.2

Multiply $2$ by $16$ .

$f (4) = 32 - 16 \cdot 4 + 27$

Step 10.2.1.3

Multiply $- 16$ by $4$ .

$f (4) = 32 - 64 + 27$

Step 10.2.2

Simplify by adding and subtracting.

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Step 10.2.2.1

Subtract $64$ from $32$ .

$f (4) = - 32 + 27$

Step 10.2.2.2

Add $- 32$ and $27$ .

$f (4) = - 5$

Step 10.2.3

The final answer is $- 5$ .

$y = - 5$

Step 11

These are the local extrema for $f (x) = 2 x^{2} - 16 x + 27$ .

$(4, - 5)$ is a local minima

Step 12