# 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?

May 26, 2015

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 $f \left(x\right) = 5 \sqrt{x} - 1$, the $5$ stretches the graph vertically and the $- 1$ translates down $1$.

It is clear that the graph is concave down on $\left(0 , \infty\right)$.
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

$f ' \left(x\right) = \frac{5}{2 \sqrt{x}}$

$f ' ' \left(x\right) = - \frac{5}{4 {\sqrt{x}}^{3}}$ is always negative on $\left(0 , \infty\right)$

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