MCQ
If $x = {t^2}$, $y = {t^3}$, then ${{{d^2}y} \over {d{x^2}}} =$
- A${3 \over 2}$
- ✓${3 \over {(4t)}}$
- C${3 \over {2(t)}}$
- D${{3t} \over 2}$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{3}{{4\sqrt x }} = \frac{3}{{4t}}$.
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$(A)$ $f$ has a local maximum at $x=2$
$(B)$ $f$ is decreasing on $(2,3)$
$(C)$ there exists some $c \in(0, \infty)$ such that $f ^{\prime \prime}( c )=0$
$(D)$ $f$ has a local minimum at $x=3$