Functions

False-True-True-False.

If \(g\left(x_{1}\right)=g\left(x_{2}\right),\) then \(x_{1}=f\left(g\left(x_{1}\right)\right)=f\left(g\left(x_{2}\right)\right)=x_{2},\) so \(g\) is one-to-one. Also \(f\) is onto because each \(x \in T\) is in the image of \(f,\) namely \(x=f(g(x)) .\) The other two statements are false, for example by constructing an \(S\) that is a larger finite set than \(T\).

Ans. B-D-D-C.
Explanation. If \(g\circ f\) is one-one, the \(f\) must be one-one. If not, there are two elements \(x\) and \(y\) with \(x\neq y\) such that \( f(x) = f(y) \), which implies that \(g(x) = g(y)\). A contradiction. However, \(f\) need not be onto and \(g\) need not be onto or one-one as the example below shows.

The first two functions have \(x=1\) and \(x=0\) as their vertical asymptotes. The denominator is always positive for the third function, so there are no vertical asymptotes for this function.

Removable discontinuities

In all the functions, the term \(x\) can be factored out from the numerator and the denominator. Hence, \(x=0\) is a removable discontinuity for all the functions.

Since \(f(0)\) is non-zero, \(f(0)=f(0)^2 \implies f(0)=1\). Since \(f(x) = f(\sqrt{x})^2 \), the range of \(f\) is non-negative. We will show that \(f(x)=1\).

Let \(p\in (0,1)\) be an arbitrary point and \(f(p) = q\). \begin{align} f(p^2) &= f(p)^2 = q^2 \\ f(p^4) &= q^4 \\ &\vdots \\ f(p^{2^n}) &= q^{2^n} \end{align} Since \(|p|< 1\) the sequence \({p^{2^n}}\) converges to 0 as \(n\rightarrow \infty\). Since the function is continuous: \[ f(0) = 1 = \lim_{n\rightarrow \infty} q^{2^n} \] The sequence \(q^{2^n}\) must converge to 1. This is possible only if \(q=1\). By continuity, \(f(1)=1\) too. Therefore, the conditions imply that \(f\) is unique and that \(f(x)=1\).

The proof is similar to the previous proof. For \(x=1\) we have: \[f(1^2) = f(1)^2 \implies f(1)(f(1)-1) = 0\] So \(f(1)\) is either 0 or 1.

Lemma. If \(f(1)=0\), then \(f(x)=0\).
Proof. For a contradiction, let us say there exists a point \(p\) such that \(f(p)=q\), where \(q>0\). \begin{align} f(\sqrt{p})^2 &= f(p) = q \\ f(\sqrt{p}) &= \sqrt{q} \\ f(p^{ 1/2^n } ) &= q^{1/2^n} \\ \lim_{n\rightarrow \infty} f(p^{ 1/2^n } ) &= q^{1/2^n} \\ f(1) &= 1 \;\;\;\; \mbox{ (a contradiction) }\;\; \square \end{align}

Lemma. If \(f(1)=1\), then \(f(x) \neq 0 \;\forall x\).
Proof. For a contradiction, let us say there exists a point \(p\) such that \(f(p)=0\). \begin{align} f(\sqrt{p})^2 &= f(p) = 0 \\ f(\sqrt{p}) &= 0 \\ f(p^{ 1/2^n }) &= 0 \;\;\;\;\mbox{ (by induction)}\\ \lim_{n\rightarrow \infty} f(p^{ 1/2^n } ) &= 0 \\ f(1) &= 0 \;\;\;\; \mbox{ (a contradiction) }\;\; \square \end{align}

For any \(p>4\), the following function satisfies the conditions: \[ f(x) = \left\{\begin{array}{lr} x& \text{for } 0 < x \leq 1\\ x^{-p} & \text{for } 1 < x < \infty \\ \end{array} \right. \]

\begin{align} \int_0^{\infty}f(x)dx &= \int_0^{1}x dx + \int_1^{\infty} x^{-p} dx \\\\ &= \left. \frac{x^2}{2} \right \rvert_{0}^{1} \; +\; \left. \frac{x^{-p+1}}{-p+1} \right \rvert_{1}^{\infty} \\\\ &= \frac{1}{2} + \frac{1}{p-1} \\\\ &< \frac{5}{6} \;\;\;\mbox{ since }\; p>4 \end{align}

Claim 1: If \(f\) is one-to-one, then it is onto.

Proof: Let \(m=n+k\). \[f^m(x) = f^n(x) \implies f^n( f^k(x) ) = f^n(x) \implies f^k(x) = x \text{ for all } x\] The second implication follows since \(f(p)=f(q)\) implies that \(p=q\) for a one-to-one function. Since \(f^k\) is an identity function, its range must be \(X\). So the range of \(f\) must also be \(X\), which proves the claim. \(\;\square\)

Claim 2: If \(f\) is onto, then it is one-to-one.

Proof: We prove by contradiction. Assume that \(f\) is onto but not one-to-one. So there must be an \(x\) such that \(f(a_1) = f(b_1) = x\) for some \(a_1\neq b_1\). Since \(f\) is onto we should be able to find two numbers \(a_2\) and \(b_2\) such that \[ f(a_2) = a_1 \text{ and } f(b_2) = b_1 \] Also, \(a_1\neq b_1 \implies a_2\neq b_2\). It follows that: \[ f^2(a_2) = x \text{ and } f^2(b_2) = x \] In general, for every natural number \(t>1\), we should be able to find two different numbers \(a_t,b_t\) with the property that \(f^t(a_t) = f^t(b_t) = x\) and \(f(a_t) = a_{t-1} \text{ and } f(b_t) = b_{t-1} \). In particular, consider the numbers \(a_m\) and \(b_m\), where \(f^m(a_m) = f^m(b_m) = x\). Due to the condition \(f^m=f^n\), we have: \[ x = f^m(a_m) = f^n(a_m) = a_{m-n} = a_k \] \[ x = f^m(b_m) = f^n(b_m) = b_{m-n} = b_k \] This implies that \( a_k=b_k \), which is a contradiction.

(a) \(\cos \left(\frac{1}{x}\right)\) is sandwiched between -1 and \(1,\) so \(\lim_{x \rightarrow 0} f(x)=0=a\) makes \(f\) continuous.

(b) Now \(f^{\prime}(0)=\lim_{h \rightarrow 0} \frac{h^{2} \cos \left(\frac{1}{h}\right)-0}{h}=\lim_{h \rightarrow 0} h \cos \left(\frac{1}{h}\right)\) which is similarly 0.

(c) For nonzero \(x,\) calculate \(f^{\prime}(x)=2 x \cos \left(\frac{1}{x}\right)+\sin \left(\frac{1}{x}\right),\) so \(\lim_{x \rightarrow 0} f^{\prime}(x)\) does not exist \(\operatorname{as} \lim_{x \rightarrow 0} 2 x \cos \left(\frac{1}{x}\right)=0\) and \(\lim_{x \rightarrow 0} \sin \left(\frac{1}{x}\right)\) does not exist.