The table shows some values of the derivative of an unknown function f.Complete the table by finding the derivative of each transformation of f, it possible a) g(x) = f(x) - 2 b) h(x) = 2 f(x) c) r(x) = f(-3x)

Step 1
The derivative properties:
$\frac{d}{dx}(f(x)-a)=f^{\prime }(x)$
$\frac{d}{dx}(af(x))=a{f}^{\prime }(x)$
calculate the derivative of $g(x)=f(x)-2$ with respect to x as follows
$g^{\prime }(x)=\frac{d}{dx}(f(x)-2)$
$=f^{\prime }(x)$

Step 2
Now calculate the derivative of $h(x)=2f(x)$ with respect to x as follows
$h^{\prime }(x)=\frac{d}{dx}=(2f(x))$
$=2f^{\prime }(x)$
$\therefore h^{\prime }(x)=2f^{\prime }(x)$

Step 3
Now calculate the derivative of $r(x)=f(-3x)$ with respect to x as follows
$r^{\prime }(x)=\frac{d}{dx}=f(-3x)$
$=-3f^{\prime }(-3x)$
As $r^{\prime }(x)=-3f^{\prime }(-3x),$ compute the the value of
$-3f^{\prime }(-3x)$
$r^{\prime }(-2)=3f^{\prime }[-3(2)]$
$=-3f^{\prime }(6)$
Here $r^{\prime }(-2)=-3f^{\prime }(6)$ cannot be computed the values
of $f^{\prime }(6)$ is not known.
$r^{\prime }(-1)=-3f^{\prime }[-3(-1)]$
$=-3f^{\prime }(3)$
$=-3(-5)$
$=15$
$r^{\prime }(0)=-3f^{\prime }[-3(0)]$
$=-3(-\frac{1}{3})$
$=1$
And $r^{\prime }(1)=-3f[-3(1)]$
$=-3f^{\prime }(-3)$
Hense $-3f^{\prime }(-3)$ can not be compluted.