# Given a f'(x), how to draw its f(x)?

## Where $f ' \left(x\right)$ is the following function: Jun 30, 2018

Because $f ' \left(x\right) > 0$ , $\forall$$x$$\in$$\left(- \infty , 1\right)$ and $f ' \left(x\right) > 0$ $\forall$$x$$\in \left(1.6 , 2\right)$

$f$ will be strictly increasing in the intervals $\left(- \infty , 1\right]$, $\left[1.6 , 2\right]$

Because $f ' \left(x\right) < 0$ , $\forall$$x$$\in$$\left(1 , 1.6\right)$

$f$ will be strictly decreasing in $\left[1 , 1.6\right]$

We can see that $f$ has a global maximum at ${x}_{0} = 1$ because
$f$ will be increasing in $\left(- \infty , 1\right]$ and decreasing in $\left[1 , 1.6\right]$ and $f ' \left(1\right) = 0$

Now all that is left so we can draw the graph of $f$ is the concavity.

We notice from the graph given that $f '$ is decreasing in $\left(- \infty , 0.25\right]$, increasing in $\left[0.25 , 0.75\right]$ , decreasing in $\left[0.75 , 1.4\right]$ and increasing in $\left[1.4 , 2\right]$ which means that $f$ will be
concave at $\left(- \infty , 0.25\right]$ and $\left[0.75 , 1.4\right]$ and convex at $\left[0.25 , 0.75\right]$, $\left[1.4 , 2\right]$

Now we have all that is needed to draw ${C}_{f}$.