Applications of Derivatives practice questions — AP Calculus BC

$f^{\mathrm{\prime }}(x)<0\iff 0<x<2$. So $f$ is decreasing on $(0,2)$. (Increasing on $(-\mathrm{\infty },0)$ and $(2,+\mathrm{\infty })$.)

$A(x)=10x-x^2$. $A^{\mathrm{\prime }}(x)=10-2x=0\Rightarrow x=5$. The maximum-area rectangle is the square $5\times 5$, with area $25$.

$a(t)=x^{\mathrm{\prime }\mathrm{\prime }}(t)=6t-12$. At $t=1$: $a=-6$ m/s².

$f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$. $f^{\mathrm{\prime }\mathrm{\prime }}<0\iff x<0$. So $f$ is concave on $(-\mathrm{\infty },0)$.

$f^{\mathrm{\prime }}(x)=2x$ and $f^{\mathrm{\prime }\mathrm{\prime }}(x)=2>0$ for all $x\in \mathbb{R}$.

Since $f^{\mathrm{\prime }\mathrm{\prime }}(x)>0$ everywhere, $f$ is convex (concave up) on all of $\mathbb{R}$ — no inflection point.

$f^{\mathrm{\prime }}(x)=3x^2-3$, $f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$. $f^{\mathrm{\prime }\mathrm{\prime }}$ changes sign at $x=0$, so the inflection is at $x=0$.

$f^{\mathrm{\prime }}(x)=-2x+4=0\Rightarrow x=2$. $f^{\mathrm{\prime }\mathrm{\prime }}(x)=-2<0$, confirming a local maximum.

$f^{\mathrm{\prime }}(x)=3x^2-3=3(x-1)(x+1)$.

$f^{\mathrm{\prime }}(x)<0$ when $(x-1)(x+1)<0$, i.e. on $(-1,1)$.

So $f$ is decreasing on $(-1,1)$.

$v(t)=3t^2-12t+9$. At $t=2$: $v=12-24+9=-3$ m/s (the particle is moving backward).

$f^{\mathrm{\prime }}(x)=3x^2-3a$, so $f^{\mathrm{\prime }}(2)=12-3a=0\Rightarrow a=4$. Check: $f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$, so $f^{\mathrm{\prime }\mathrm{\prime }}(2)=12>0$, confirming a local minimum.