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?

Question

4404 views around the world

You can reuse this answer
Creative Commons License

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.