How do you sketch the curve $y=(x+5)^(1/4)$ by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

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Answer 1 · 2017-06-28T00:03:12

The function $f(x) = (x+5)^(1/4)$ is defined and continuous for $(x+5) > 0$ , that is for $x in [-5,+oo]$ .

At the boundaries of the domain we have:

$lim_(x->5^+) (x+5)^(1/4) = 0$

$lim_(x->+oo) (x+5)^(1/4) = oo$

inside the domain the function has only positive values.

Evaluate the derivatives:

$f'(x) = 1/4(x+5)^(-3/4)$

$f''(x) = -3/8(x+5)^(-7/4)$

So we have that:

$f'(x) > 0$ for $x in (-5,+oo)$ and $lim_(x->5^+) f'(x) = oo$

$f''(x) < 0$ for $x in (-5,+oo)$

We can then conclude that $f(x)$ is strictly increasing and has no relative maxima or minima and is concave down in its domain:

graph{(x+5)^(1/4) [-10, 10, -5, 5]}

How do you sketch the curve y=(x+5)^(1/4)y=(x+5)^(1/4) by finding local maximum, minimum, inflection points, asymptotes, and intercepts?