Injectivity, Surjectivity, and Bijectivity for Functions on $[-1,1]$

We consider $A=B=\{x\mid -1\le x\le 1\}$. For each function $f:A\to B$, we determine where $f$ is injective , surjective , or bijective.
Key idea: on an interval, injectivity is guaranteed if the function is strictly monotone; surjectivity depends on whether the range equals $[-1,1]$.
We will treat “where it is injective/surjective” as the largest subdomain(s) of $A$ on which those properties hold.

Mermaid roadmap:

Common toolkit (used in both parts)

• Image (range): $\mathrm{Im}(f)=\{f(x)\mid x\in A\}$
• Injective: $\forall x_1\ne x_2,\ f(x_1)\ne f(x_2)$
• Surjective onto $B$: $\mathrm{Im}(f)=B$
• Bijective: both injective and surjective

Injectivity subdomains: if a function changes monotonicity at some point (e.g., where derivative changes sign), it is typically injective only on intervals that do not cross that point.

Surjectivity: if $\max f(A)<1$ or $\min f(A)>-1$, then it cannot be onto $[-1,1]$.

Important note (for this problem): Both parts specify $A=B=[-1,1]$, so “surjective” means onto $[-1,1]$ exactly.

(i) $g(x)=\sin(\pi x)$ on $A=[-1,1]$

We analyze $\sin(\pi x)$ on the symmetric interval $[-1,1]$.

Step 1: Determine injective regions

Consider where $\sin(\pi x)$ achieves its “turning point” on $[-1,1]$.

• $\sin(\pi x)$ has derivative $g'(x)=\pi\cos(\pi x)$.
• $g'(x)=0$ when $\cos(\pi x)=0$, i.e. $\pi x=\frac{\pi}{2}+k\pi \Rightarrow x=\frac12+k$. Within $[-1,1]$, the critical point is at $x=\frac12$ and also at $x=-\frac12$ (from $\pi x=-\frac{\pi}{2}$).

Thus the monotonicity switches at $x=-\frac12$ and $x=\frac12$. Therefore $g$ is:

• strictly increasing on $[-1,-\frac12]$,
• strictly decreasing on $[-\frac12,\frac12]$,
• strictly increasing on $[\frac12,1]$.

Hence $g$ is injective on each of those subintervals, but not injective on subdomains that include both sides of a switch (e.g., including both $-1$ and $0$ typically creates collisions because the function is not one-to-one globally).

We can also exhibit a collision explicitly:

• $g(-x)=\sin(-\pi x)=-\sin(\pi x)$ (odd), so collisions occur when two different $x$ give the same sine value—e.g. symmetry around $\pm\frac12$ can produce equal outputs at distinct points within $[-1,1]$.

Largest injective subdomains (maximal intervals):

• $[-1,-\frac12]$,
• $[-\frac12,\frac12]$,
• $[\frac12,1]$.

(Each is an interval where $g$ is strictly monotone, hence injective.)

Step 2: Determine surjectivity onto $B=[-1,1]$

Compute the range over $[-1,1]$:

• $g(-1)=\sin(-\pi) = 0$
• $g(-\tfrac12)=\sin(-\tfrac{\pi}{2})=-1$
• $g(0)=\sin(0)=0$
• $g(\tfrac12)=\sin(\tfrac{\pi}{2})=1$
• $g(1)=\sin(\pi)=0$

Since $g$ is continuous and it attains $-1$ and $1$ on $[-1,1]$, its image is exactly $[-1,1]$.

Therefore, $g$ is surjective on $A$.

Step 3: Bijectivity

For bijectivity, we need injective on all of $A$ plus surjective.

• We already know $g$ is not injective on all of $[-1,1]$ (because it is not monotone on the whole interval).
• It is surjective on all of $[-1,1]$.

So $g$ is not bijective on $A$, but it is bijective on each maximal injective subinterval where the range matches the codomain of that restricted mapping.

However, the codomain is still $B=[-1,1]$ in the original statement. If we interpret the question strictly as: “where is $g$ injective/surjective/bijective as a function $A\to B$,” then:

• Surjectivity on $A$ holds.
• Injectivity holds only on those subdomains (as restricted functions), while surjectivity generally fails on smaller subdomains unless their ranges still cover $[-1,1]$.

So the final classification:

• Injective: on $[-1,-\tfrac12]$, $[-\tfrac12,\tfrac12]$, $[\tfrac12,1]$
• Surjective: on all $[-1,1]$
• Bijective: none on all of $[-1,1]$ (not injective globally)

(ii) $h(x)=\dfrac{2x}{3}$ on $A=[-1,1]$

This is a linear function, so its behavior is straightforward.

Step 1: Injectivity on $A$

A linear function $h(x)=ax$ (with $a\ne 0$) is strictly monotone (here increasing because $a=\tfrac{2}{3}>0$), hence injective on all of $[-1,1]$.

So $h$ is injective on $[-1,1]$.

Step 2: Surjectivity onto $B=[-1,1]$

Compute the image of the endpoints:

• $h(-1)=-\tfrac{2}{3}$
• $h(1)=\tfrac{2}{3}$

Because $h$ is continuous and linear, $\mathrm{Im}(h)=[-\tfrac{2}{3},\tfrac{2}{3}]$.

But $B=[-1,1]$, and $[-\tfrac{2}{3},\tfrac{2}{3}]\ne [-1,1]$. In particular, values like $1$ and $-1$ are not attained.

So $h$ is not surjective on $[-1,1]$.

Step 3: Bijectivity

Bijectivity requires both injectivity and surjectivity onto the given codomain $B$.

• Injective: yes on all of $[-1,1]$
• Surjective: no onto $[-1,1]$

Therefore $h$ is not bijective (as a map $A\to B$).

If we were allowed to change the codomain to $\left[-\tfrac{2}{3},\tfrac{2}{3}\right]$, then $h$ would become bijective—but under the problem’s fixed $B=[-1,1]$, it is not.

Final answers (summary tables)

(i) $g(x)=\sin(\pi x)$ with $A=B=[-1,1]$

• Injective on: $[-1,-\tfrac12]$, $[-\tfrac12,\tfrac12]$, $[\tfrac12,1]$
• Surjective on: $[-1,1]$
• Bijective on: nowhere on all of $[-1,1]$ (not injective globally)

(ii) $h(x)=\dfrac{2x}{3}$ with $A=B=[-1,1]$

• Injective on: $[-1,1]$
• Surjective on: nowhere (range $[-\tfrac23,\tfrac23]$ misses $\pm 1$)
• Bijective on: nowhere (injective but not surjective onto $[-1,1]$)