$f\text{'}\left(1\right)=-2$; $g\text{'}\left(1\right)=0,5$; $h\text{'}\left(1\right)=-4$; $k\text{'}\left(1\right)=0$

Use your graphing calculator to check your graph.

Find the x-axis intersects of the chosen function.
$f$: max. $f\left(0\right)=4$
$g$: min. $g\left(0\right)=\sqrt{\left(3\right)}$
$h$: no extrema
$k$: max. $k\left(1\right)=3$

$\langle -1,1\rangle$

$x=1$

No, to do so you would need to know the rule of the original function $f$.

The correct graph would be that of function $f\left(x\right)=-\frac{2}{3}{x}^3+2x+2$, but you were (probably) not able to deduce that from the information.
Your answer is good enough if the graph goes through the point $(0,2)$ and has a maximum at $x=1$ and a minimum at $x=-1$.

Use your graphing calculator to check your graph.

$v\left(t\right)=3,2t$

$3,2t\approx 22,22$; solving this for $t$ gives $t\approx 6,9$ s

Own answer.

$40$

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Use your graphing calculator to do so.

Find x-axis intersects of the slope function and enter the corresponding $x$ -values in the function rule for $f$.
You should find: min. $f(2,53)\approx -26,26$ and max. $f(-0,52)\approx 2,26$.