Exercises — Applications — increasing - decreasing, local extrema (first derivative test)

What we do: plug the test values into $f^{\prime }$ and read the sign. We do not need $f$ itself — the sign of $f^{\prime }$ is the entire map of rising/falling.

• $f^{\prime }(0)=(0-2)(0+3)=(-2)(3)=-6<0$ → decreasing at $x=0$.
• $f^{\prime }(5)=(5-2)(5+3)=(3)(8)=24>0$ → increasing at $x=5$.

Two ways to be critical: $f^{\prime }=0$ or $f^{\prime }$ undefined while $x$ is in the domain of $f$.

• $f^{\prime }=0$: the fraction is zero only when its top is zero → $x-1=0\Rightarrow x=1$. ✓
• $f^{\prime }$ undefined: at $x=4$ the denominator is zero. But $x=4$ is not in the domain of $f$, so it is not a critical point of $f$ — a point must belong to the domain to count.

Answer: the only critical point is $x=1$.

Step 1 — differentiate. $f^{\prime }(x)=2x-6$ (power rule term by term; the constant $+5$ has slope $0$). Step 2 — critical points. $2x-6=0\Rightarrow x=3$. $f^{\prime }$ is a polynomial, defined everywhere, so no "undefined" points. Step 3–4 — test points either side of $x=3$:

• $x=0$: $f^{\prime }(0)=-6<0$ → decreasing on $(-\infty ,3)$.
• $x=4$: $f^{\prime }(4)=2>0$ → increasing on $(3,\infty )$. Step 5 — classify. Sign goes $-\to +$ → local minimum at $x=3$. Its value is $f(3)=9-18+5=-4$. Answer: decreasing on $(-\infty ,3)$, increasing on $(3,\infty )$; local min at $(3,-4)$.

Step 1. $f^{\prime }(x)=6x^2-6x-12$. Step 2. Set to zero: $6x^2-6x-12=0\Rightarrow x^2-x-2=0\Rightarrow (x-2)(x+1)=0\Rightarrow x=-1,2$. Step 3–4. Three intervals; test one point each (this is exactly the $+,-,+$ chart in the top figure):

• $x=-2$: $f^{\prime }=6(4)-6(-2)-12=24+12-12=24>0$ → increasing on $(-\infty ,-1)$.
• $x=0$: $f^{\prime }=-12<0$ → decreasing on $(-1,2)$.
• $x=3$: $f^{\prime }=6(9)-18-12=54-30=24>0$ → increasing on $(2,\infty )$. Step 5.
• At $x=-1$: $+\to -$ → local max, value $f(-1)=2(-1)-3(1)-12(-1)+1=-2-3+12+1=8$.
• At $x=2$: $-\to +$ → local min, value $f(2)=2(8)-3(4)-24+1=16-12-24+1=-19$. Answer: local max $(-1,8)$, local min $(2,-19)$.

Step 1. $f^{\prime }(x)=\frac{2}{3}{x}^{-1/3}=\frac{2}{3\sqrt[3]{x}}$. Step 2 — critical points. $f^{\prime }=0$ has no solution (the numerator $2/3$ is never zero). But $f^{\prime }$ is undefined at $x=0$ (division by $\sqrt[3]{0}=0$), and $x=0$ is in the domain of $f$ (since $0^{2/3}=0$). So $x=0$ is a critical point — the "derivative-does-not-exist" kind. Step 3–4. Test either side. The key is the sign of $\sqrt[3]{x}$: it is negative for $x<0$ and positive for $x>0$.

• $x=-1$: $\sqrt[3]{-1}=-1$, so $f^{\prime }=\frac{2}{3(-1)}=-\frac{2}{3}<0$ → decreasing on $(-\infty ,0)$.
• $x=1$: $\sqrt[3]{1}=1$, so $f^{\prime }=\frac{2}{3}>0$ → increasing on $(0,\infty )$. Step 5. Sign $-\to +$ → local minimum at $x=0$, value $f(0)=0$. Geometrically this is a sharp cusp — the graph comes down, points, and goes back up. Answer: local min at $(0,0)$; no other critical points.

Part A — $f(x)=x^4$ (the honest minimum). $f^{\prime }(x)=4x^3$. Setting $4x^3=0$ gives the single critical point $x=0$. Test either side:

• $x=-1$: $f^{\prime }=4(-1{)}^3=-4<0$ → decreasing.
• $x=1$: $f^{\prime }=4(1{)}^3=4>0$ → increasing. Sign $-\to +$ → local minimum at $(0,0)$. So $x^4$ genuinely has an extremum — it is not a counterexample to "zero slope ⇒ extremum," it just happens to be a case where the implication holds.

Part B — $g(x)=(x-1{)}^5$ (the real trap). $g^{\prime }(x)=5(x-1{)}^4$ (chain rule: outer power $5$, inner derivative $1$). Setting it to zero gives $x=1$, one critical point. Since $(x-1{)}^4\ge 0$ always (even power), $g^{\prime }\ge 0$ everywhere:

• $x=0$: $g^{\prime }=5(-1{)}^4=5>0$ → increasing.
• $x=2$: $g^{\prime }=5(1{)}^4=5>0$ → increasing. Sign $+\to +$ — same sign on both sides → no extremum. The point $(1,0)$ is an inflection with a horizontal tangent. Answer: $x^4$ has a local min at $0$; $g=(x-1{)}^5$ has a critical point at $x=1$ but no extremum. The difference: $f^{\prime }=4x^3$ does change sign at $0$, while $g^{\prime }=5(x-1{)}^4$ (even power) does not.

Domain first. We need $4-x^2\ge 0\Rightarrow -2\le x\le 2$. So the domain is the closed interval $[-2,2]$. Step 1 — differentiate (product + chain rule). Write $f=x(4-x^2{)}^{1/2}$. $f^{\prime }(x)=1\cdot (4-x^2{)}^{1/2}+x\cdot \frac{1}{2}(4-x^2{)}^{-1/2}(-2x)=\sqrt{4-x^2}-\frac{x^2}{\sqrt{4-x^2}}.$ Combine over one denominator: $f^{\prime }(x)=\frac{(4-x^2)-x^2}{\sqrt{4-x^2}}=\frac{4-2x^2}{\sqrt{4-x^2}}.$ Step 2 — critical points.

• $f^{\prime }=0$: numerator zero → $4-2x^2=0\Rightarrow x^2=2\Rightarrow x=\pm \sqrt{2}$. Both lie inside $(-2,2)$. ✓
• $f^{\prime }$ undefined: denominator zero at $x=\pm 2$ — but these are the endpoints of the domain, so we classify them separately with one-sided reasoning below. Step 3–4 — sign of $f^{\prime }$ (denominator $\sqrt{4-x^2}>0$ on the interior, so the sign follows the numerator $4-2x^2$):
• $x=-1.5$: $4-2(2.25)=-0.5<0$ → decreasing on $(-2,-\sqrt{2})$.
• $x=0$: $4-0=4>0$ → increasing on $(-\sqrt{2},\sqrt{2})$.
• $x=1.5$: $-0.5<0$ → decreasing on $(\sqrt{2},2)$. Step 5 — classify interior critical points.
• At $x=-\sqrt{2}$: $-\to +$ → local min. Value $f(-\sqrt{2})=-\sqrt{2}\cdot \sqrt{4-2}=-\sqrt{2}\cdot \sqrt{2}=-2$.
• At $x=\sqrt{2}$: $+\to -$ → local max. Value $f(\sqrt{2})=\sqrt{2}\cdot \sqrt{2}=2$. Step 6 — classify the endpoints (one-sided test). On a closed domain an endpoint is a local extremum if $f$ moves the "downhill" way as you step inward from it.
• At the left endpoint $x=-2$: just to the right, on $(-2,-\sqrt{2})$, $f^{\prime }<0$ (decreasing). So stepping inward from $x=-2$ the function goes down, meaning $x=-2$ sits above its immediate neighbours → local maximum. Value $f(-2)=-2\cdot \sqrt{4-4}=-2\cdot 0=0$.
• At the right endpoint $x=2$: just to the left, on $(\sqrt{2},2)$, $f^{\prime }<0$, so approaching $x=2$ the function is still falling; that makes $x=2$ lower than its immediate neighbours → local minimum. Value $f(2)=2\cdot 0=0$. Answer: interior local min $(-\sqrt{2},-2)$ and local max $(\sqrt{2},2)$; endpoint local max $(-2,0)$ and endpoint local min $(2,0)$.

Domain. $x^2+1$ is never zero → domain is all real $x$. Step 1 — quotient rule. $f^{\prime }(x)=\frac{(1)(x^2+1)-x(2x)}{(x^2+1{)}^2}=\frac{x^2+1-2x^2}{(x^2+1{)}^2}=\frac{1-x^2}{(x^2+1{)}^2}.$ Step 2. Denominator $(x^2+1{)}^2>0$ always, so $f^{\prime }=0\iff 1-x^2=0\Rightarrow x=\pm 1$. No undefined points. Step 3–4 — sign follows numerator $1-x^2$:

• $x=-2$: $1-4=-3<0$ → decreasing on $(-\infty ,-1)$.
• $x=0$: $1>0$ → increasing on $(-1,1)$.
• $x=2$: $-3<0$ → decreasing on $(1,\infty )$. Step 5.
• At $x=-1$: $-\to +$ → local min, value $f(-1)=\frac{-1}{2}=-\frac{1}{2}$.
• At $x=1$: $+\to -$ → local max, value $f(1)=\frac{1}{2}=\frac{1}{2}$. Answer: local min $(-1,-\frac{1}{2})$, local max $(1,\frac{1}{2})$.

Step 1. $f^{\prime }(x)=3x^2+a$. Reasoning. A local extremum needs a critical point where $f^{\prime }$ changes sign. Since $f^{\prime }=3x^2+a$ is an upward parabola in $x$:

• If $a>0$: $3x^2+a>0$ for all $x$ (never zero) → $f^{\prime }$ never zero, $f$ strictly increasing → no extrema. ✓
• If $a=0$: $f^{\prime }=3x^2\ge 0$, zero only at $x=0$ but $+\to +$ (no sign change) → no extrema (inflection). ✓
• If $a<0$: $3x^2+a=0$ has two roots $x=\pm \sqrt{-a/3}$, and the parabola dips below zero between them → $f^{\prime }$ goes $+,-,+$ → genuine local max then local min → has extrema. Answer: $f$ has no local extrema exactly when $a\ge 0$.

Goal. Show: for any $a<b$ in $I$, $f(a)<f(b)$ (the definition of increasing). Setup. Since $f$ is differentiable on $I$ (that's what "$f^{\prime }(x)$ exists" means), it is also continuous on $[a,b]$ and differentiable on $(a,b)$ — exactly the hypotheses of the Mean Value Theorem. Apply MVT. There exists some $c\in (a,b)$ with $f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}.$ Why MVT and not something else? We need to convert a statement about the slope $f^{\prime }$ into a statement about values $f(b)$ vs $f(a)$. MVT is the exact bridge: it equates the average slope over $[a,b]$ to an actual instantaneous slope somewhere inside. Finish. Rearrange: $f(b)-f(a)=f^{\prime }(c)(b-a)$. By hypothesis $f^{\prime }(c)>0$, and $b-a>0$ since $a<b$. A positive times a positive is positive, so $f(b)-f(a)>0$, i.e. $f(b)>f(a)$. As $a<b$ were arbitrary, $f$ is increasing on $I$. $■$

Step 1. $f^{\prime }(x)=\cos x-\sin x$. Step 2 — interior critical points. $\cos x=\sin x\Rightarrow \tan x=1$. Before dividing by $\cos x$, check the points where $\cos x=0$ (namely $x=\frac{\pi }{2},\frac{3\pi }{2}$) are not lost:

• at $x=\frac{\pi }{2}$: $f^{\prime }=\cos \frac{\pi }{2}-\sin \frac{\pi }{2}=0-1=-1\ne 0$.
• at $x=\frac{3\pi }{2}$: $f^{\prime }=\cos \frac{3\pi }{2}-\sin \frac{3\pi }{2}=0-(-1)=1\ne 0$. Neither is critical, so dividing was safe. On $[0,2\pi ]$, $\tan x=1$ gives $x=\frac{\pi }{4}$ and $x=\frac{5\pi }{4}$. Step 3–4 — sign of $f^{\prime }=\cos x-\sin x$:
• $x=0$: $1-0=1>0$ → increasing on $(0,\frac{\pi }{4})$.
• $x=\pi$: $-1-0=-1<0$ → decreasing on $(\frac{\pi }{4},\frac{5\pi }{4})$.
• $x=2\pi$: $1-0=1>0$ → increasing on $(\frac{5\pi }{4},2\pi )$. Step 5 — interior critical points.
• At $x=\frac{\pi }{4}$: $+\to -$ → local max, value $\sin \frac{\pi }{4}+\cos \frac{\pi }{4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$.
• At $x=\frac{5\pi }{4}$: $-\to +$ → local min, value $-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}=-\sqrt{2}$. Step 6 — endpoints (one-sided test).
• Left endpoint $x=0$: just to the right, on $(0,\frac{\pi }{4})$, $f^{\prime }>0$ (increasing). Stepping inward the function rises, so $x=0$ is below its neighbours → local minimum, value $f(0)=\sin 0+\cos 0=0+1=1$.
• Right endpoint $x=2\pi$: just to the left, on $(\frac{5\pi }{4},2\pi )$, $f^{\prime }>0$ (increasing), so approaching $2\pi$ the function is still rising; that makes $x=2\pi$ above its neighbours → local maximum, value $f(2\pi )=\sin 2\pi +\cos 2\pi =0+1=1$. Answer: interior local max $\left(\frac{\pi }{4},\sqrt{2}\right)$, interior local min $\left(\frac{5\pi }{4},-\sqrt{2}\right)$; endpoint local min $(0,1)$ and endpoint local max $(2\pi ,1)$.