Attempting to draw G(x) from G'(x)

Question

We know $G(0) = 0$

Okay, so I have the above graph but I'm having a difficult time translating it into the graph of $G(x)$.

What I know so far is that the slope changes abruptly from 0 to 2 at $x=0$. I also know that the slope gets extremely close to -1 but is never -1 itself, and it gets really close to 0 but isn't 0 itself. Finally, I know that $G(x)$ has a negative slope for $x<0$ and a positive slope for $x>0$.

What I don't understand is how we can show that the slope is getting really close to -1 or 0 on $G(x)$?

Answer 1

A common application of this kind of question is when you have the speed $v=\frac {ds}{dt}$ which corresponds to $G'(x)$ and you want to find the displacement $s $ which corresponds to $G(x)$. In that situation it helps to remember that the area under the velocity-time curve is equal to the displacement. For you that means that the area under the $G'(x)$ curve is equal to the change in $G(x)$.

If you like, $G(1)=G(0) + \int_0^1 G'(x) dx$

You can estimate the areas from your graph.

$G(1)\approx G(0)+1.5$

$G(2)\approx G(1)+0.75$

$G(3)\approx G(2)+0.3$

$G(4)\approx G(3)+0.1$

$G(5)\approx G(4)+0$

These give you a set of points that you can try to join up in a "dot-to-dot" fashion. You also know that for small positive $x$ the gradient is about 2.

You can also work backwards, so that ...

$G(0)\approx G(-1) -0.25 \Rightarrow G(-1) \approx G(0)+0.25$

Answer 2

Another way to attack this problem is to draw a tangent field diagram. This is usually done when the gradient is a function of both $x$ and $y$, but there is no reason why not to go ahead in this instance.

The basic idea is to draw a short line with gradient $G'(x)$ at every point $(x,y)$:

You then take your starting point $(0,0)$ and draw a curve that would fit the general pattern.