Quiz 3 (Solution)

Date: MATH 1210-4 - Spring 2004

1. Find the natural domains:
   1. $Dom(\frac{3}{x^2+7})$.

      The only problem that $\frac{3}{x^2+7}$ could have is a 0 in the denominator. But $x^2+7$ is at least $7$, so is never 0, and the function has no problem. Therefore $Dom(\frac{3}{x^2+7}) = \ensuremath{\mathbb{R}}$.

   2. $Dom(\frac{3}{x+7})$.

      The only problem that $\frac{3}{x+7}$ could have is a 0 in the denominator. Then, $x+7=0 \ensuremath{\Rightarrow}\xspace x=-7$, so the only problem that the function has is at $x=-7$. Therefore $Dom(\frac{3}{x+7}) = \ensuremath{\mathbb{R}} -\{-7\}$.

   3. $Dom(\sqrt{3+x})$.

      This function has problems when the numbers inside the root are negative. So, if $3+x \geq 0$ (better, $x\geq -3$), there are no problems. Thus $Dom(\sqrt{3+x}) = [-3,+\infty)$.

2. Let $f(x)=x^5+x$, $g(x)=\frac{1}{x}$, and $h = f\circ g$.
   1. Write an explicit formula for $h(x)$.

      $$h(x) = (f \circ g)(x) = f(g(x)) = f(\frac{1}{x}) = (\frac{1}{x})^5 + \frac{1}{x} = \frac{1}{x^5} + \frac{1}{x}.$$

      
      

   2. Compute $h(2)$.

      $$h(2) = \frac{1}{2^5} + \frac{1}{2} = \frac{1}{32} + \frac{1}{2} = \frac{1+16}{32} = \frac{17}{32}.$$

      
      

3. Evaluate, without using a calculator, $\tan(\frac{\pi}{3})$.

   First, we remember that $\tan(x) = \frac{\sin(x)}{\cos(x)}$. Then, since we know that $\sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$ and $\cos(\frac{\pi}{3}) = \frac{1}{2}$ we conclude

   $$\tan(\frac{\pi}{3}) = \frac{\sin(\frac{\pi}{3})}{\cos(\frac{\pi}{3})} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}.$$

   
   

4. Specify if $f(x) = \frac{x}{x^2-1}$ is even, odd, or neither. Prove your claim.

   We have to find if there is a relation between $f(-x)$ and $f(x)$.

   $$f(-x) = \frac{-x}{(-x)^2-1} = \frac{-x}{x^2-1} = - \frac{x}{x^2-1} = -f(x),$$

   
   
   so that $f$ is an odd function.