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)$.

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

    2. Compute $ h(2)$.

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

  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

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

  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)$.

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

    so that $ f$ is an odd function.



Javier Fernandez 2004-02-06