Continuity and the Intermediate Value Theorem

1. $f(3)=5$ is defined.

2. For $x\ne3$, $\dfrac{x^2-9}{x-3}=\dfrac{(x-3)(x+3)}{x-3}=x+3$, so

$\lim_{x\to3}f(x)=\lim_{x\to3}(x+3)=6,$

which exists and is finite.

3. $6\ne5$: the limit and the value disagree, so $f$ is not continuous at $3$.

Because the limit exists, the discontinuity is removable: setting $f(3)=6$ (equivalently, using $x+3$ for every $x$) makes $f$ continuous on all of $(-\infty,\infty)$.

For $x>3$, $|x-3|=x-3$ so $f(x)=1$; for $x<3$, $|x-3|=-(x-3)$ so $f(x)=-1$. Hence

$\lim_{x\to3^-}f(x)=-1,\qquad \lim_{x\to3^+}f(x)=1.$

$f(3)=1$ equals the right-hand limit, so $f$ is right-continuous at $3$; it differs from the left-hand limit, so $f$ is not left-continuous, hence not continuous. Both one-sided limits are finite and unequal: a jump of size $1-(-1)=2$. No choice of $f(3)$ can repair it — one value can match only one side.

$4-x^2$ is a polynomial, continuous everywhere, and $\ln$ is continuous on $(0,\infty)$; by the composition theorem $\ln(4-x^2)$ is continuous wherever $4-x^2>0$, i.e. on $(-2,2)$.

$x+1$ is continuous and vanishes only at $x=-1$. By the quotient rule for continuity, $g$ is continuous at every $x$ in $(-2,2)$ with $x\ne-1$:

$(-2,-1)\cup(-1,2).$

At $x=-1$ the numerator tends to $\ln3>0$ while $x+1\to0$: as $x\to-1^-$, $x+1\to0^-$ so $g\to-\infty$; as $x\to-1^+$, $g\to+\infty$. An infinite discontinuity, not removable. ($x=\pm2$ are domain endpoints, not discontinuities — $g$ is undefined on a whole side of each.)

Each piece is a polynomial, so only $x=-1$ and $x=2$ can fail.

At $x=-1$: $f(-1)=-a+b=\lim_{x\to-1^+}f(x)$, while $\lim_{x\to-1^-}f(x)=2(-1)+1=-1$. So $-a+b=-1$.

At $x=2$: $f(2)=2a+b=\lim_{x\to2^-}f(x)$, while $\lim_{x\to2^+}f(x)=4-3=1$. So $2a+b=1$.

Subtract: $3a=2$, so $a=\frac23$ and $b=-1+a=-\frac13$.

Check: $-\frac23-\frac13=-1$ ✓ and $\frac43-\frac13=1$ ✓.

(a) Let $g(x)=\cos x-x$, continuous everywhere as a difference of continuous functions, so continuous on $\left[0,\tfrac{\pi}{2}\right]$.

$g(0)=1>0,\qquad g\!\left(\tfrac{\pi}{2}\right)=0-\tfrac{\pi}{2}<0.$

$N=0$ lies strictly between, so by the IVT there is $c$ in $\left(0,\tfrac{\pi}{2}\right)$ with $g(c)=0$, i.e. $\cos c=c$.

(b) The IVT concludes "at least one $c$": $h(x)=(x-1)(x-2)(x-3)$ has $h(0)<0<h(4)$ and three zeros in $(0,4)$. Uniqueness here: on $\left[0,\tfrac{\pi}{2}\right]$ both $\cos x$ and $-x$ are strictly decreasing, so $g$ is strictly decreasing and takes the value $0$ at most once. At least one and at most one: exactly one ($c\approx0.739$).

Let $g(x)=f(x)-x$, continuous on $[0,1]$. Since $0\le f\le1$,

$g(0)=f(0)\ge0,\qquad g(1)=f(1)-1\le0.$

If $g(0)=0$ or $g(1)=0$, that endpoint is a fixed point. Otherwise $g(0)>0>g(1)$ and the IVT gives $c$ in $(0,1)$ with $g(c)=0$, i.e. $f(c)=c$. (Strictness matters: the IVT needs $N=0$ strictly between the endpoint values.)

On $(0,1)$ take $h(x)=\frac{x+1}{2}$: continuous, and $0<x<1$ gives $\frac12<h(x)<1$, so $h$ maps $(0,1)$ into itself. But $h(x)=x$ forces $x=1$, which is not in $(0,1)$. The lost hypothesis is the closed interval — the sign change happens only at the endpoint that $(0,1)$ leaves out.