Quiz 1 (Solution)


Date: MATH 1210-4 - Spring 2004

  1. Find an equation for a line with slope $ 3$ and containing the point $ (-1,1)$.

    Using the point-slope form for the line, $ y-y_0 = m (x-x_0)$, we have

    $\displaystyle y-1 = 3 (x-(-1)) = 3(x+1) \ensuremath{\Rightarrow}\xspace y-1 = 3(x+1).$    

    Equivalently, solving for $ y$ we obtain, $ y=3x+4$. Both equations are equivalent.

  2. Let $ f(x) = 5x^2-x$.
    1. Write down $ \frac{f(x+h)-f(x)}{h}$, but do not simplify anything.

      $\displaystyle \frac{f(x+h)-f(x)}{h} = \frac{5(x+h)^2-(x+h)-(5x^2-x)}{h}$    

    2. Now simplify your previous expression as much as you can.

      \begin{displaymath}\begin{split}\frac{5(x+h)^2-(x+h)-(5x^2-x)}{h} &= \frac{5(x^2...
...h^2 -h}{h} \\ &= \frac{h(10x+5h -1)}{h} = 10x+5h-1. \end{split}\end{displaymath}    

    3. Using your previous computations find $ f'(x) =
\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}$.

      $\displaystyle f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0} (10x+5h-1) = 10x-1.$    

      That is, $ f'(x)=10x-1$.

  3. For each of the following functions find the derivative (use properties of the derivative).
    1. $ f(x)=3x^{1234}$.

      $\displaystyle (3x^{1234})' = 3(x^{1234})' = 3 \cdot 1234\cdot x^{1234-1} = 3702 x^{1233}.$    

    2. $ g(x)=-x^9+33x^3+\frac{1}{3}$.

      \begin{displaymath}\begin{split}(-x^9+33x^3+\frac{1}{3})' &= (-x^9)'+(33x^3)'+(\...
...0 \\ &= -9x^8 + 33\cdot 3\cdot x^2 = -9x^8 + 99 x^2 \end{split}\end{displaymath}    



Javier Fernandez 2004-01-26