Exercises — Intermediate Value Theorem

Continuity: $x^2$ is a polynomial, so continuous everywhere — in particular on the closed interval $[0,3]$. ✓ Endpoint heights: $f(0)=0$ and $f(3)=9$. Guaranteed values: every $N$ with $\min (0,9)<N<\max (0,9)$, i.e. every $N$ in the open interval $(0,9)$. So $N=5$ lies in $(0,9)$ ⇒ hit (by IVT there is a $c\in (0,3)$ with $c^2=5$, namely $c=\sqrt{5}\approx 2.236$). $N=10$ is not between $0$ and $9$ (it is above $\max =9$), so IVT says nothing — and indeed $x^2\le 9$ on $[0,3]$, so $10$ is not hit. ✓

(a) YES. Sign change ($-3<0<4$) and continuous ⇒ IVT gives a root. (b) IVT-SILENT. Both positive, so $0$ is not between $2$ and $6$. IVT gives no guarantee (there might still be a root, but not by this theorem). (c) IVT-SILENT. Continuity fails, so the hypothesis is broken and IVT applies to nothing here — it makes no promise either way. (A jump can skip over $0$, so we cannot even conclude a root exists.)

Continuous: polynomial. ✓ Endpoints: $f(1)=1-1-1=-1<0$, $f(2)=8-2-1=5>0$. This is a sign change ⇒ IVT gives $c\in (1,2)$ with $f(c)=0$. ✓ Bisection step: midpoint $m=1.5$. $f(1.5)=3.375-1.5-1=0.875>0$. The sign change now lives between $x=1$ (negative) and $x=1.5$ (positive), so the root is in $[1,1.5]$. See Bisection Method.

Set $g(x)=\cos x-x$, continuous (cosine and a line are continuous). Why this move? "Solve $\cos x=x$" is a root question in disguise. $g(0)=\cos 0-0=1>0$. $g(1)=\cos 1-1\approx 0.5403-1=-0.4597<0$. Sign change ⇒ IVT gives $c\in (0,1)$ with $g(c)=0$, i.e. $\cos c=c$. ✓ (This $c\approx 0.739$ is the famous "Dottie number".)

Build $g(x)=f(x)-N=2^x-3x-(-1)=2^x-3x+1$, continuous. $g(0)=1-0+1=2>0$. $g(1)=2-3+1=0$. The right endpoint is exactly $0$! So $c=1$ works, but $1$ is an endpoint. To land strictly inside, check a mid value: $g(0.5)=\sqrt{2}-1.5+1=\sqrt{2}-0.5\approx 0.914>0$, still positive — the only zero here is at the boundary. So $f(1)=2-3=-1=N$ exactly. The value $N=-1$ is attained, at the endpoint. ✓ Lesson: IVT needs $N$ strictly between the endpoint values to force an interior $c$. Here $N=f(1)$ exactly, so we get the value trivially at the boundary rather than via IVT's interior guarantee.

Let $p(x)=a_n{x}^n+\cdots +a_0$ with $n$ odd and $a_n\ne 0$. Assume $a_n>0$ without loss of generality (WLOG — meaning we lose nothing by picking this case, since if instead $a_n<0$ we simply argue about $-p$, which has the same roots and now a positive leading coefficient). Continuity: polynomials are continuous everywhere. ✓ End behaviour: for large $|x|$ the leading term $a_n{x}^n$ dominates. Because $n$ is odd, $x^n\to +\infty$ as $x\to +\infty$ and $x^n\to -\infty$ as $x\to -\infty$. So there exist a large $b$ with $p(b)>0$ and a large-negative $a$ with $p(a)<0$. IVT: on $[a,b]$, $p$ is continuous and $p(a)<0<p(b)$ (a sign change) ⇒ some $c\in (a,b)$ with $p(c)=0$. $■$ Why odd? For even degree both ends go the same way, so no forced sign change (e.g. $x^2+1$ has no real root).

Recall "$N$ between $f(0)$ and $f(2)$" means $\min (f(0),f(2))<N<\max (f(0),f(2))$. Here both endpoint heights are $1$, so $\min =\max =1$ and the condition $1<N<1$ is impossible — the open interval of guaranteed values is empty. IVT is therefore silent about every value except $1$ itself.

• $1.5$: not guaranteed. Counterexample: $f(x)=1$ constant never reaches $1.5$.
• $0.5$: not guaranteed either. Same constant example.
• But IVT does not forbid them: a continuous $f$ starting and ending at $1$ could easily dip to $0.5$ or rise to $1.5$ in between. IVT gives a sufficient condition, never a necessary one. Equal endpoints ⇒ no forced intermediate values, but other values may still occur.

Define $g(x)=f(x)-x$, continuous (difference of continuous functions). Why? Turn "fixed point $f(x)=x$" into "root $g(x)=0$". See Fixed Point Theorems.

• $g(0)=f(0)-0=f(0)\ge 0$ (outputs land in $[0,1]$, so $f(0)\ge 0$).
• $g(1)=f(1)-1\le 0$ (since $f(1)\le 1$). Cases: if $g(0)=0$ then $c=0$; if $g(1)=0$ then $c=1$. Otherwise $g(0)>0>g(1)$: a sign change, IVT gives $c\in (0,1)$ with $g(c)=0$, i.e. $f(c)=c$. $■$ Codomain $[0,2]$: now $f(1)\le 2$, so $g(1)=f(1)-1$ could be positive (e.g. $f\equiv 2$ gives $g(0)=2>0$, $g(1)=1>0$, no sign change) — no fixed point guaranteed. The argument needed outputs to stay inside the same interval as the inputs.

Define $h(\theta )=T(\theta )-T(\theta +180^{\circ })$, continuous (difference of continuous). Why? "Equal temperatures" becomes "$h=0$". Evaluate at $\theta =0$ and $\theta =180^{\circ }$: $h(0)=T(0)-T(180^{\circ }),h(180^{\circ })=T(180^{\circ })-T(360^{\circ })=T(180^{\circ })-T(0)=-h(0).$ So $h(180^{\circ })$ is the exact negative of $h(0)$.

• If $h(0)=0$: done, $\theta =0$ works.
• Else $h(0)$ and $h(180^{\circ })$ have opposite signs (a sign change) ⇒ IVT on $[0^{\circ },180^{\circ }]$ gives ${\theta }^{\ast }$ with $h({\theta }^{\ast })=0$, i.e. $T({\theta }^{\ast })=T({\theta }^{\ast }+180^{\circ })$. $■$

Existence: from 2.1, a root exists in $(1,2)$. Uniqueness via the derivative. The derivative is $f^{\prime }(x)=3x^2-1$. Set $r=\frac{1}{\sqrt{3}}\approx 0.577$; then $f^{\prime }(x)>0$ exactly when $x^2>\frac{1}{3}$, i.e. on $(-\infty ,-r)$ and on $(r,\infty )$, and $f^{\prime }(x)<0$ on the middle band $(-r,r)$. So $f$ is strictly increasing on $(-\infty ,-r)$, strictly decreasing on $(-r,r)$, and strictly increasing on $(r,\infty )$.

• A strictly monotonic piece can cross any horizontal level at most once, so $f$ has at most one root on each of the three pieces — at most three total. We now kill two of them.
• Local max at $x=-r$: $f(-r)={\left(-\frac{1}{\sqrt{3}}\right)}^3+\frac{1}{\sqrt{3}}-1=\frac{2}{3\sqrt{3}}-1\approx 0.385-1=-0.615<0$. Since $-r$ is the highest point of the left-and-middle hump, all of $(-\infty ,r]$ sits at or below this local max value $<0$. Being strictly negative there, $f$ has no root on $(-\infty ,r]$.
• That leaves only the last strictly-increasing piece $(r,\infty )$, where $f$ rises from $f(r)<0$ through $0$ once. Strict monotonicity forbids a second crossing. Hence exactly one real root. $■$ Moral: IVT (existence) + monotonicity from $f^{\prime }$ (no second crossing) = "exactly one". IVT alone can never say "one". See Rolle's Theorem.

Look at value $N=2$.

• On $[0,1]$: $f(0)=4$, $f(1)=0$, and $2$ satisfies $\min (0,4)<2<\max (0,4)$, i.e. $0<2<4$. Continuous ⇒ IVT gives $c_1\in (0,1)$ with $f(c_1)=2$.
• On $[1,2]$: $f(1)=0$, $f(2)=4$, again $0<2<4$. IVT gives $c_2\in (1,2)$ with $f(c_2)=2$. Since $c_1<1<c_2$, they are distinct. So $f=2$ at least twice. $■$

No contradiction: IVT demands continuity on the whole closed interval. $g(x)=1/x$ is not defined, hence not continuous, at $x=0\in [-1,1]$. The hypothesis fails, so IVT makes no promise — and indeed $g$ leaps from $-\infty$ to $+\infty$, skipping $0$. Repair (both sides of the blow-up): on any closed interval that avoids $0$, $g$ is continuous, so IVT applies there. There are two natural families — one on each side of the discontinuity.

• Positive side, e.g. $[1,2]$: $g$ is continuous here (no $0$ inside), $g(1)=1$ and $g(2)=0.5$. IVT guarantees every value in the open interval $(0.5,1)$ is hit — for instance $g(4/3)=0.75$ lands right in that range. ✓
• Negative side, e.g. $[-1,-0.5]$: $g$ is continuous here too, $g(-1)=-1$ and $g(-0.5)=-2$. IVT guarantees every value in the open interval $(-2,-1)$ is hit — for instance $g(-4/3)=-0.75$ lands in that range. ✓ Neither interval crosses $0$, so $g$ never needs to attain $0$; there is no contradiction. More generally, IVT applies to $1/x$ on any closed interval lying entirely inside $(0,\infty )$ or entirely inside $(-\infty ,0)$ — just never on an interval straddling $0$. IVT was never wrong — it was simply never invited on $[-1,1]$ because its continuity ticket was invalid there.

IVT gives existence or uniqueness? ::: Existence only — pair with monotonicity/Rolle for uniqueness. To solve $f(x)=N$, what function do you build? ::: $g(x)=f(x)-N$, then find a sign change. Define "sign change" of $g$ on $[a,b]$. ::: $g(a)$ and $g(b)$ have opposite signs, i.e. $g(a)\cdot g(b)<0$. Define "$N$ is between $p$ and $q$". ::: $\min (p,q)<N<\max (p,q)$ (empty when $p=q$). Why must odd-degree polynomials have a real root? ::: Opposite end behaviours give a sign change; IVT forces a crossing. Equal endpoint values — what does IVT guarantee? ::: Nothing beyond that value; it is silent (but does not forbid) other values. Why is $1/x$ on $[-1,1]$ not a counterexample? ::: It is discontinuous at $0$, so the hypothesis fails and IVT never applies.