Function flash cards

1. 1.Find the domain of $f(x)=\sqrt{\log_{0.5}\left(\dfrac{x-1}{x+1}\right)}$.

   Need $\dfrac{x-1}{x+1}>0 \Rightarrow x<-1$ or $x>1$. Also need $\log_{0.5}\left(\frac{x-1}{x+1}\right)\ge 0$; since base $0.5<1$, this means $0<\dfrac{x-1}{x+1}\le 1$. Solving $\dfrac{x-1}{x+1}\le 1 \Rightarrow \dfrac{-2}{x+1}\le 0 \Rightarrow x>-1$. Intersecting with $x<-1$ or $x>1$ gives domain $x\in(1,\infty)$.

   Hint: Log defined only for positive argument; since base $<1$, $\log_{0.5}t\ge0 \iff 0

2. 2.If $f(x)$ is an odd function and $g(x)$ is an even function, what is the nature of $h(x)=f(x)\cdot g(x)$? Also state nature of $f(g(x))$.

   $h(x)=f(x)g(x)$ is odd since $h(-x)=f(-x)g(-x)=(-f(x))(g(x))=-h(x)$. Also $f(g(x))$ is even since $g(-x)=g(x)\Rightarrow f(g(-x))=f(g(x))$.

   Hint: Odd×Even=Odd; f(even function) is always even.

3. 3.A function $f:\mathbb{R}\to\mathbb{R}$ satisfies $f(x+y)=f(x)+f(y)$ for all $x,y$ and $f$ is continuous. If $f(1)=2$, find $f(x)$.

   $f(x)=2x$. Cauchy's functional equation with continuity forces $f(x)=kx$; using $f(1)=2$ gives $k=2$.

   Hint: Cauchy equation + continuity $\Rightarrow$ linear function.

4. 4.If $f(x)=\dfrac{x}{\sqrt{1+x^2}}$, find $f(f(x))$ and identify the pattern for $f_n(x)$ ($n$-fold composition).

   $f(f(x))=\dfrac{x}{\sqrt{1+2x^2}}$. In general, $f_n(x)=\dfrac{x}{\sqrt{1+nx^2}}$, provable by induction.

   Hint: Substitute $f(x)$ into itself and simplify the radical.

5. 5.Let $f(x)=x^2-2x$ for $x\ge 1$. Find $f^{-1}(x)$ and its domain.

   Writing $y=x^2-2x=(x-1)^2-1 \Rightarrow (x-1)^2=y+1 \Rightarrow x=1+\sqrt{y+1}$ (taking $+$ root since $x\ge1$). So $f^{-1}(x)=1+\sqrt{x+1}$, with domain $x\ge -1$ (range of $f$).

   Hint: Complete the square; choose the branch consistent with $x\ge1$.

6. 6.Find the domain of $f(x)=\sqrt{\log_{0.5}(x^2-x-2)}$.

   Since the base is less than $1$, $\log_{0.5}t\ge 0 \iff 00\Rightarrow x<-1$ or $x>2$; the right part gives $x^2-x-3\le 0\Rightarrow \dfrac{1-\sqrt{13}}{2}\le x\le\dfrac{1+\sqrt{13}}{2}$. Intersecting, the domain is $\left[\dfrac{1-\sqrt{13}}{2},-1\right)\cup\left(2,\dfrac{1+\sqrt{13}}{2}\right]$.

   Hint: Use $\log_{0.5}t\ge0 \iff 0

7. 7.Find the domain of $f(x)=\sqrt{\sin^{-1}(2x)}+\dfrac{1}{\sqrt{x^2-1}}$.

   For the first term we need $-1\le 2x\le1$ (domain of $\sin^{-1}$) and $\sin^{-1}(2x)\ge0$ (for the square root), which together force $0\le x\le \dfrac12$. For the second term we need $x^2-1>0$, i.e. $x<-1$ or $x>1$. These two requirements have no common value of $x$, so the domain of $f$ is the empty set $\varnothing$.

   Hint: Work out each piece's domain separately before intersecting them.

8. 8.Find the domain of $f(x)=\dfrac{1}{\sqrt{[x]-x}}$, where $[\cdot]$ is the greatest integer function.

   For every real $x$, $[x]\le x$, so $[x]-x\le 0$ always, with equality exactly when $x$ is an integer. The expression $[x]-x$ is therefore never strictly positive, so $\sqrt{[x]-x}$ is never a positive real number and $f$ is undefined for every real $x$; its domain is $\varnothing$.

   Hint: Recall $[x]\le x$ always — can $[x]-x$ ever be positive?

9. 9.Find the domain of $f(x)=\log_{x}(x^2-5x+6)$ (variable base).

   A logarithm's base must satisfy $x>0,\ x\ne1$. The argument must be positive: $x^2-5x+6>0\Rightarrow (x-2)(x-3)>0\Rightarrow x<2$ or $x>3$. Combining both conditions gives the domain $(0,1)\cup(1,2)\cup(3,\infty)$.

   Hint: Don't forget the base restrictions $x>0$ and $x\ne1$ as well as the argument condition.

10. 10.Find the domain of $f(x)=\sqrt{3-x}+\dfrac{1}{\sqrt{x^2-1}}$.

    The first term needs $3-x\ge0\Rightarrow x\le3$. The second needs $x^2-1>0\Rightarrow x<-1$ or $x>1$. Intersecting these gives the domain $(-\infty,-1)\cup(1,3]$.

    Hint: Find each term's domain separately, then intersect.

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