# Thread: Given the following graph of h(x), determine/sketch graphs of h'(x) and h''(x)

1. ## Given the following graph of h(x), determine/sketch graphs of h'(x) and h''(x)

Okay so... I know h'(x) is going to be a parabola. I think it will also bounce off the x-axis at x=2... but I'm not sure. A little help?

2. ## Re: Given the following graph of h(x), determine/sketch graphs of h'(x) and h''(x)

Originally Posted by porkchop

Okay so... I know h'(x) is going to be a parabola. I think it will also bounce off the x-axis at x=2... but I'm not sure. A little help?
yes, the vertex of h' will be at (2,0)

3. ## Re: Given the following graph of h(x), determine/sketch graphs of h'(x) and h''(x)

Yes. Looking at the gradient of h(x) it is positive and decreasing, is 0 when x=2 and then is positive and increasing. (It will look like a parabola but I dont think we can definitely say it is one)
Now consider in the same way the gradient of this graph.

4. ## Re: Given the following graph of h(x), determine/sketch graphs of h'(x) and h''(x)

So, will h''(x) be a decreasing line then?

5. ## Re: Given the following graph of h(x), determine/sketch graphs of h'(x) and h''(x)

Originally Posted by porkchop
So, will h''(x) be a decreasing line then?
increasing

6. ## Re: Given the following graph of h(x), determine/sketch graphs of h'(x) and h''(x)

You cannot assume that h is a cubic. There exist polynomials of all odd powers, greater than or equal to 3, with graphs that look like that. You can argue that the graph of the first derivative comes down, with positive y values, to 0 at x= 2, then increases again, but not necessarily a parabola. You can then look at that graph to argue the second derivative is negative for x= 0, becoming 0 at x= 2, then becomes positive. But it is not necessarily a straight line.