#### Answer

(b), (d) and (f)

#### Work Step by Step

$f^{(k)}(x)=0$ implies that the derivative eventually becomes zero for all $k$ values.
a) As there is a $x^{-1}$ term, by the power rule, the power of $x$ becomes more and more negative when we differentiate again and again. Therefore, the power can never be 0.
b) $f'(x)=3x^{2}$
$f''(x)=6x$
$f^{(3)}(x)=6x^{0}=6$
$f^{(4)}(x)=0$
Differentiating $0$ again and again, we get 0. Therefore, the condition is satisfied.
c) While differentiating $\sqrt x=x^{1/2}$ again and again, as the power of $x$ never becomes 0, $f^{(k})(x)$ will never be equal to 0.
d) The power of $x$ will become $6-6=0$ after differentiating 6 times. If we differentiate after that, we get $f^{(7)}(x)=0$, etc. Therefore, the condition is satisfied.
e) While differentiating $x^{9/5}$ again and again, as the power of $x$ never becomes 0, $f^{(k})(x)$ will never be equal to 0.
f) $f'(x)= 4x+15x^{4}$
$f''(x)=4+60x^{3}$
$f'''(x)=0+180x^{2}$
$f^{(4)}(x)=360x$
$f^{(5)}(x)=360$
$f^{(6)}(x)=0$
The condition is satisfied.