Find the local and extreme values of the function f(x) = \sqrt{x^2 + 6x + 18}

Question:

Find the local and extreme values of the function {eq}f(x) = \sqrt{x^2 + 6x + 18} {/eq}

Extrema:

{eq}\\ {/eq}

In order to find local extrema,

1. Differntiate f(x) and equate it with 0 to get critical points. (f'(x) =0)

2. Now, Substitute it in function again.

We will get our required extrema.

Answer and Explanation:

{eq}\\ {/eq}

We have,

{eq}f(x) = \sqrt{x^2 + 6x + 18} {/eq}

Now,

{eq}f'(x) = \frac{1}{2\sqrt{x^2 + 6x + 18}} (2x + 6) \\ \text{ for f'(x) = 0}...

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