Analytical Applications of Differentiation

Let $f(x)=x^2$ on $[1,3]$. It is a polynomial, so continuous and differentiable everywhere. The average rate is

Set $f'(c)=2c=4$, so $c=2$, which lies in $(1,3)$. ✓

Let $f(x)=x^2-2x$ on $[0,2]$. Then $f(0)=0=f(2)$, and $f$ is a polynomial, so Rolle applies. Solving $f'(c)=2c-2=0$ gives $c=1$ in $(0,2)$. ✓

$f'(x)=3x^2-3=3(x-1)(x+1)$, so critical numbers are $x=-1,\,1$. The sign chart (above) tests $f'$ once in each interval: $f'(-2)=9>0$, $f'(0)=-3<0$, $f'(2)=9>0$.

Local max at $x=-1$ with $f(-1)=2$; local min at $x=1$ with $f(1)=-2$.

$f''(x)=6x$. Then $f''<0$ for $x<0$ (concave down) and $f''>0$ for $x>0$ (concave up); the concavity changes at $x=0$, so $(0,0)$ is an inflection point. Using the Second Derivative Test at the critical numbers: $f''(-1)=-6<0$ (local max) and $f''(1)=6>0$ (local min), matching the First Derivative Test.

For $f(x)=x^4$, $f''(x)=12x^2$, which is $0$ at $x=0$ but never changes sign ($f''\ge 0$). So $x=0$ is not an inflection point—$f$ is concave up throughout.

$f'(x)=2x-4=0\Rightarrow x=2$ (in the interval). Evaluate all candidates:

Absolute maximum $3$ at $x=0$; absolute minimum $-1$ at $x=2$.

Cut squares of side $x$ from each corner and fold. The volume is

Then $V'(x)=(12-2x)^2+x\cdot 2(12-2x)(-2)=(12-2x)(12-6x)$. Setting $V'=0$: $x=6$ (gives $V=0$) or $x=2$. Since $V'$ changes $+\to-$ at $x=2$, it is a maximum:

Combining the work above: increasing on $(-\infty,-1)$ and $(1,\infty)$, decreasing on $(-1,1)$; local max $(-1,2)$, local min $(1,-2)$; concave down for $x<0$, up for $x>0$; inflection $(0,0)$.

For $f(x)=\tfrac13x^3-x$ we get $f'(x)=x^2-1$ and $f''(x)=2x$. Notice: $f$ has a local max where $f'$ crosses zero going $+\to-$ ($x=-1$); $f$ has an inflection where $f''$ crosses zero and $f'$ bottoms out ($x=0$).

Show $|\sin a-\sin b|\le|a-b|$ for all $a,b$. Apply the MVT to $g(x)=\sin x$ on $[a,b]$: there is a $c$ with

Taking absolute values, $\dfrac{|\sin b-\sin a|}{|b-a|}=|\cos c|\le 1$, so $|\sin a-\sin b|\le|a-b|$.

Suppose the graph shown is $f'$ (not $f$), with $f'(x)=x^2-1$. Then $f$ is increasing where $f'>0$: on $(-\infty,-1)$ and $(1,\infty)$; decreasing on $(-1,1)$. So $f$ has a local max at $x=-1$ and a local min at $x=1$. Since $f'$ is decreasing for $x<0$ and increasing for $x>0$, $f$ is concave down then concave up, with an inflection point at $x=0$.

For $f(x)=x^4-4x^3$, $f''(x)=12x^2-24x=12x(x-2)$. Sign of $f''$: positive for $x<0$, negative on $(0,2)$, positive for $x>2$. Two sign changes give two inflection points, at $x=0$ and $x=2$. Contrast with $f(x)=x^4$, where $f''=12x^2\ge0$ never changes sign—no inflection despite $f''(0)=0$.

Minimize the surface area of a closed cylinder of fixed volume $V$. Constraint: $\pi r^2 h=V\Rightarrow h=\dfrac{V}{\pi r^2}$. Objective:

Then $S'(r)=4\pi r-\dfrac{2V}{r^2}=0\Rightarrow r^3=\dfrac{V}{2\pi}\Rightarrow r=\sqrt[3]{\dfrac{V}{2\pi}}$. Since $S''=4\pi+\dfrac{4V}{r^3}>0$, this is a minimum. Substituting back gives $h=2r$: the most economical can is as tall as it is wide.

Closest point on $y=\tfrac1x$ ($x>0$) to the origin. Minimize $D^2=x^2+\dfrac{1}{x^2}$ (minimizing $D^2$ minimizes $D$). Then $\dfrac{d(D^2)}{dx}=2x-\dfrac{2}{x^3}=0\Rightarrow x^4=1\Rightarrow x=1$. The point is $(1,1)$ with distance $\sqrt2$.