Calculus Examples

$g (x) = \sqrt{x^{2} - 16 x + 64}$

Step 1

Find the first derivative of the function.

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Step 1.1

Rewrite $\frac{d}{d x} [\sqrt{x^{2} - 16 x + 64}]$ as $\frac{d}{d x} [x - 8]$ .

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Step 1.1.1

Factor using the perfect square rule.

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Step 1.1.1.1

Rewrite $64$ as $8^{2}$ .

$\frac{d}{d x} [\sqrt{x^{2} - 16 x + 8^{2}}]$

Step 1.1.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$16 x = 2 \cdot x \cdot 8$

Step 1.1.1.3

Rewrite the polynomial.

$\frac{d}{d x} [\sqrt{x^{2} - 2 \cdot x \cdot 8 + 8^{2}}]$

Step 1.1.1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 8$ .

$\frac{d}{d x} [\sqrt{{(x - 8)}^{2}}]$

Step 1.1.2

Pull terms out from under the radical, assuming positive real numbers.

$\frac{d}{d x} [x - 8]$

Step 1.2

By the Sum Rule, the derivative of $x - 8$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 8]$ .

$\frac{d}{d x} [x] + \frac{d}{d x} [- 8]$

Step 1.3

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

$1 + \frac{d}{d x} [- 8]$

Step 1.4

Since $- 8$ is constant with respect to $x$ , the derivative of $- 8$ with respect to $x$ is $0$ .

$1 + 0$

Step 1.5

Add $1$ and $0$ .

$1$

Step 2

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

$g'' (x) = 0$

Step 3

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

$1 = 0$

Step 4

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

No Local Extrema

Step 5

No Local Extrema

Step 6