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

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

   
   
   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.

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

      
      

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

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

      
      

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

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

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

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

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

      
      

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

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