Unit 5: Analytical Applications of Differentiation

For $f(x)=x^2$ on $[1,3]$, the polynomial is continuous on $[1,3]$ and differentiable on $(1,3)$. The average rate is $$\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4.$$ Since $f'(x)=2x$, solve $2c=4$ to get $c=2$, which lies in $(1,3)$.

For $f(x)=x^2-2x$ on $[0,3]$, continuity guarantees both global extrema. Since $f'(x)=2x-2$, the only interior critical point is $x=1$. Compare $f(0)=0$, $f(1)=-1$, and $f(3)=3$. Thus the global minimum is $-1$ at $x=1$, while the global maximum is $3$ at the endpoint $x=3$.

For $f(x)=x^3-3x$, $$f'(x)=3x^2-3=3(x-1)(x+1).$$ The derivative is zero at $x=-1$ and $x=1$, so these values split the real line into three intervals.

For $f(x)=x^3-3x$, $$f'(x)=3(x-1)(x+1),$$ so the critical points are $x=-1$ and $x=1$. The derivative signs are positive on $(-\infty,-1)$, negative on $(-1,1)$, and positive on $(1,\infty)$. Therefore $f'$ changes $+\to-$ at $x=-1$, giving a local maximum, and $-\to+$ at $x=1$, giving a local minimum.

For $f(x)=x^2+2x-3$ on $[-3,2]$, $f'(x)=2x+2=0$ gives the interior critical point $x=-1$. The candidates are therefore $-3$, $-1$, and $2$.

For $f(x)=x^3-3x^2$, $$f''(x)=6x-6=6(x-1).$$ This is negative for $x<1$ and positive for $x>1$, so $f$ is concave down on $(-\infty,1)$ and concave up on $(1,\infty)$. The concavity changes at $x=1$, and $f(1)=-2$, so $(1,-2)$ is an inflection point.

For $f(x)=x^2+4x+1$, $f'(x)=2x+4$, so the only critical point is $c=-2$. Since $f''(x)=2>0$, $f$ has a local minimum at $x=-2$. The polynomial is continuous on $(-\infty,\infty)$ and has only this one critical point, so the local minimum is also the absolute minimum; its value is $f(-2)=-3$.

Suppose $f'$ is positive on $(-\infty,-2)$, negative on $(-2,1)$, and positive on $(1,\infty)$. Then $f$ has a local maximum at $x=-2$ and a local minimum at $x=1$. If $f''<0$ for $x<0$ and $f''>0$ for $x>0$, the sketch is concave down before $0$, concave up after $0$, and has an inflection point at $x=0$ if that point lies on the graph.

For $f(x)=x^3-3x$, $$f'(x)=3x^2-3,\qquad f''(x)=6x.$$ The cubic has horizontal tangents at $x=-1$ and $x=1$, exactly where the derivative parabola crosses the $x$-axis. The parabola decreases for $x<0$ and increases for $x>0$, matching the negative and positive sides of the line $f''=6x$. The cubic therefore changes concavity at $x=0$.

For a rectangle with fixed perimeter $P>0$, let its sides be $x$ and $y$. The constraint $2x+2y=P$ gives $y=P/2-x$. Therefore $$A(x)=x\left(\frac{P}{2}-x\right),\qquad 0\le x\le\frac{P}{2}.$$ Differentiate: $A'(x)=P/2-2x$, so the interior candidate is $x=P/4$. Then $y=P/4$. The endpoint areas are $0$, while $A(P/4)=P^2/16$, so the maximum area is $P^2/16$ square units and occurs for a square.

For the fixed-perimeter rectangle from the previous method, the calculation gives side lengths $x=y=P/4$ and area $A=P^2/16$. The interpretation is: among all rectangles with perimeter $P$, the square with side length $P/4$ has the greatest possible area, $P^2/16$. If $P$ is measured in meters, the side lengths are in meters and the maximum area is in square meters.

For $x^2+4y^2=4$, implicit differentiation gives $dy/dx=-x/(4y)$. The derivative is zero when $x=0$; the original relation then gives $(0,1)$ and $(0,-1)$. It is undefined when $y=0$; the relation gives $(2,0)$ and $(-2,0)$. Thus all four are critical points of the relation; the first pair has horizontal tangents and the second pair has vertical tangents.

On the upper semicircle, $y>0$, so $d^2y/dx^2<0$: the entire upper branch is concave down. There, $dy/dx=-x/y$ is positive when $x<0$ and negative when $x>0$, so the branch increases to $(0,5)$ and then decreases. On the lower branch, $y<0$, so the concavity conclusion reverses.