$f$ is differentiable, $f^{\prime }(2)=0$, $f^{\prime }(x)>0$ on $(0,2)$ and $f^{\prime }(x)<0$ on $(2,5)$. Student: because $f^{\prime }(2)=0$, $f$ has a relative maximum at $x=2$. Judge.

$f$ is differentiable, $f^{\prime }(-1)=0$, $f^{\prime }(x)<0$ on $(-3,-1)$ and $f^{\prime }(x)>0$ on $(-1,2)$. Student: since $f^{\prime }$ changes from negative to positive, $f$ has a relative minimum at $x=-1$. Judge.

$f$ is differentiable, $f^{\prime }(4)=0$, and $f^{\prime }(x)<0$ on both $(1,4)$ and $(4,7)$. Student: $f$ has a relative minimum at $x=4$ because $f^{\prime }(4)=0$. Judge.

$f$ is continuous at $x=1$ and differentiable nearby except possibly at $1$. Also $f^{\prime }(x)<0$ on $(-2,1)$ and $f^{\prime }(x)>0$ on $(1,4)$. Student: no relative extremum occurs because $f^{\prime }(1)$ is not stated. Judge.

$f$ is differentiable, $f^{\prime }(0)=0$, $f^{\prime }(x)>0$ on $(-3,0)$ and $f^{\prime }(x)<0$ on $(0,3)$. Student: $f$ has a relative minimum at $x=0$ because $f^{\prime }$ changes sign. Judge.

$f$ is continuous, $f^{\prime \prime }(x)<0$ on $(1,3)$, and $f^{\prime \prime }(x)>0$ on $(3,6)$. Student: $f$ has an inflection point at $x=3$ because concavity changes from down to up. Judge.

$f^{\prime \prime }(1)=0$, and $f^{\prime \prime }(x)>0$ on both $(0,1)$ and $(1,2)$. Student: $f$ has an inflection point at $x=1$ because $f^{\prime \prime }(1)=0$. Judge.

$f$ is continuous at $0$, $f^{\prime \prime }(x)<0$ on $(-2,0)$, and $f^{\prime \prime }(x)>0$ on $(0,2)$. Student: there is an inflection point at $x=0$ because concavity changes sign there. Judge.

$f^{\prime }(x)<0$ on $(0,2)$ and $f^{\prime }(x)>0$ on $(2,4)$. Student: $f$ has an inflection point at $x=2$ because $f^{\prime }$ changes sign there. Judge.

A continuous $f$ on $[0,4]$ has only critical points $x=1,3$. Values: $f(0)=2$, $f(1)=7$, $f(3)=5$, $f(4)=1$. Student: the absolute maximum is $7$ by comparing all candidates. Judge.

On $[0,5]$, continuous $f$ has critical points $x=2$ and $x=4$, with $f(0)=8$, $f(2)=9$, $f(4)=3$, $f(5)=12$. Student: the absolute maximum is $9$ because $f^{\prime }$ changes from positive to negative at $2$. Judge.

On $[-1,3]$, continuous $f$ has critical points $x=0,2$. Values: $f(-1)=4$, $f(0)=-2$, $f(2)=1$, $f(3)=0$. Student: the absolute minimum is $0$ at $x=3$ because it is the smaller endpoint value. Judge.

$h$ is continuous on $[1,5]$, differentiable on $(1,5)$, $h(1)=4$, $h(5)=16$. Student: since $\frac{16-4}{5-1}=3$, the Mean Value Theorem gives $c\in (1,5)$ with $h^{\prime }(c)=3$. Judge.

$k(0)=1$ and $k(4)=7$. Student: the Intermediate Value Theorem guarantees a $c\in (0,4)$ with $k(c)=5$. No continuity fact is given. Judge.

$p$ is continuous on $[-2,4]$. Student: $p$ has an absolute maximum and an absolute minimum on $[-2,4]$ by the Extreme Value Theorem. Judge.

$r$ is continuous on $[0,2]$, differentiable on $(0,2)$, and $r(0)=r(2)=5$. Student: Rolle's Theorem gives a $c\in (0,2)$ with $r(c)=0$. Judge.

On $(0,6)$, $f^{\prime }(x)>0$ for every $x$. Student: therefore $f$ is increasing on $(0,6)$. Judge.

Student: $f$ is decreasing on $(0,5)$ because $f(0)=9$ and $f(5)=2$. No other facts are given. Judge.

On $(1,4)$, $f^{\prime \prime }(x)<0$ for all $x$. Student: thus $f$ is decreasing on $(1,4)$. Judge.

Let $g(x)={\int }_1^{x}f(t)dt$, with continuous $f$ and $f(x)<0$ on $[1,5]$. Student: by the Fundamental Theorem, $g^{\prime }(x)=f(x)<0$, so $g$ is decreasing. Judge.

Let $g(x)={\int }_0^{x}f(t)dt$, and suppose $f$ is increasing on $[0,4]$. Student: since $f$ is increasing, $g$ is increasing on $[0,4]$. Judge.

Let $g(x)={\int }_2^{x}f(t)dt$. Continuous $f$ is positive on $(2,3)$, negative on $(3,5)$, and $f(3)=0$. Student: since $g^{\prime }=f$ changes positive to negative, $g$ has a relative maximum at $x=3$. Judge.

Student: ${\sum }_{n=1}^{\infty }(-1{)}^n\frac{n+1}{n}$ converges by the Alternating Series Test because the signs alternate. Judge the justification.

Student: for $a_n=\frac{1}{n^2}$, ${\sum }_{n=1}^{\infty }{a}_n$ converges by the Ratio Test because ${\lim }_{n\to \infty }\frac{a_{n+1}}{a_n}=1$. Judge the justification.