How do you find the intervals on which the graph of $f(x)=5sqrtx-1$ is concave up or is concave down, and find the x - coordinates o the points of inflection?

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Answer 1 · 2015-05-26T18:24:57

Unless required, I would not use calculus for this.

The graph of the square root function is:

graph{y=sqrtx [-1.34, 11.146, -1.744, 4.5]}

In $f(x)=5sqrtx-1$ , the $5$ stretches the graph vertically and the $-1$ translates down $1$ .

It is clear that the graph is concave down on $(0,oo)$ .
Since the concavity does not change, there are no inflection points.

If I am required to use calculus:

Note that the domain of $f$ is #[0

$f'(x) = 5/(2sqrtx)$

$f''(x) = -5/(4sqrtx^3)$ is always negative on $(0,oo)$

Therefore, the graph of $f$ is concave down.

How do you find the intervals on which the graph of f(x)=5sqrtx-1f(x)=5sqrtx-1 is concave up or is concave down, and find the x - coordinates o the points of inflection?