SOLUTION OF EXAM 1 MATH 131 Fall 2002

1. a) (10) Evaluate the limit
   
   $$\lim_{x\rightarrow 2}\frac{2x^2-5x+2}{x^3-4x}$$
   
   
   by an algebraic procedure. Show all your algebraic steps.

   
   

   $$\lim_{x\rightarrow 2}\frac{2x^2-5x+2}{x^3-4x}= \lim_{x\right... ...tarrow 2}\frac{2x-1}{x(x+2)}=\frac{2(2)-1}{2(2+2)}=\frac{3}{8}$$
   
   

   b) (5) Support your result in part a) with a calculator. Describe your procedure and show as much data as possible.

   x	y
   1.9	.377868
   1.99	.375310
   1.9999	.375003
   1.999999	.375000
   2.1	.371661
   2.01	.374685
   2.0001	.374997
   2.000001	.374999

   As the values of $x$ approach 2 either from the right or from the left (but not equal to 2) the values of $y$ approach $.375=3/8$.

2. Given the function
   
   $$f(x) \ = \ \left\{ \begin{array}{ccc} x^2+1 & \mbox{for} & x\leq 1 \\ ax+b & \mbox{for} & x > 1, \end{array}\right.$$
   
   
   where $a$, $b$ are constants.

   a) (5) Find ${\displaystyle \lim_{x\rightarrow 1^{-}}f(x)}$ and ${\displaystyle \lim_{x\rightarrow 1^{+}}f(x)}$.

   
   

   $$\lim_{x\rightarrow 1^{-}}f(x)=\lim_{x\rightarrow 1^{-}}x^2+1=(1)^2+1=2,$$
   
   

   
   

   $$\lim_{x\rightarrow 1^{+}}f(x)=\lim_{x\rightarrow 1^{+}}ax+b=a(1)+b=a+b.$$
   
   

   b) (5) What conditions do $a$ and $b$ have to satisfy, so that $f(x)$ is continuous at $x=1$.

   
   

   For $f(x)$ to be continuous at $x=1$ we need the limit ${\displaystyle \lim_{x\rightarrow 1}f(x)}$ exist and equal $f(1)$. Since $f(1)=2$ and ${\displaystyle \lim_{x\rightarrow 1}f(x)}$ exists if and only if ${\displaystyle \lim_{x\rightarrow 1^{-}}f(x)}={\displaystyle \lim_{x\rightarrow 1^{+}}f(x)}$ we see that the condition on $a$ and $b$ is $a+b=2$.

   
   

   c) (5) Find $a$ and $b$, such that $f(x)$ is everywhere differentiable.

   
   

   First $f(x)$ must be continuous everywhere, thus $b=2-a$ from part b). Notice that $f(x)$ is differentiable at each $x\neq 1$. The derivative at $x=1$ is the limit ${\displaystyle \lim_{x\rightarrow 1}\frac{f(x)-f(1)}{x-1}}$. It exists if and only if the left hand and the right hand derivative limits are equal. We have

   
   

   $$\lim_{x\rightarrow 1^{-}}\frac{f(x)-f(1)}{x-1}=\lim_{x\right... ...{x-1}=\lim_{x\rightarrow 1^{-}}\frac{(x-1)(x+1)}{x-1}=(1)+1=2,$$
   
   

   
   

   $$\lim_{x\rightarrow 1^{+}}\frac{f(x)-f(1)}{x-1}=\lim_{x\right... ...x+(2-a)-2}{x-1}=\lim_{x\rightarrow 1^{+}}\frac{a(x-1)}{x-1}=a,$$
   
   
   Therefore $a=2$ and $b=0$.

3. a) (10) Find all horizontal and vertical asymptotes for the function
   
   $$f(x) \ = \ \frac{3x^2+1}{(2x+3)(x-1)}$$
   
   
   Explain your reasons.

   
   

   The vertical asymptotes are the lines $x=a$ such that the denominator is zero at $x=a$, but the numerator is not. We have $(2x+3)(x-1)=0$, i.e. $x=-3/2$ or $x=1$. Since $3(-3/2)^2+1=31/4\neq 0$ and $3(1)^2+1=4\neq 0$ the vertical asymptotes are $x=-3/2$ and $x=1$.

   The horizontal asymptotes are the lines $y=b$ where $b$ is the limit of $f(x)$ as $x$ approaches positive or negative infinity. We have

   
   

   $$\lim_{x\rightarrow +\infty}\frac{3x^2+1}{(2x+3)(x-1)}=\lim_{... ...c{3+1/x^2}{(2+3/x)(1-1/x)}=\frac{3+0}{(2+0)(1-0)}=\frac{3}{2}.$$
   
   

   
   

   $$\lim_{x\rightarrow -\infty}\frac{3x^2+1}{(2x+3)(x-1)}=\lim_{... ...c{3+1/x^2}{(2+3/x)(1-1/x)}=\frac{3+0}{(2+0)(1-0)}=\frac{3}{2}.$$
   
   

   Therefore the only horizontal asymptote is $y=3/2$.

   
   

   b) (5) Graph the function in part a) indicating all horizontal and vertical asymptotes.

4. (10) Determine the derivative $f'(x)$ of the given function using the definition of the derivative as a limit
   
   $$f(x) \ = \ \frac{x+1}{x-1}$$
   
   

   One can first write $f(x)=\frac{x+1}{x-1}=\frac{x-1+2}{x-1}=1+\frac{2}{x-1}$. Then

   
   

   $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{1+\frac{2}{x+h-1}-1-\frac{2}{x-1}}{h}$$
   
   
   
   
   $$=\lim_{h\to 0}\frac{2((x-1)-(x+h-1))}{(x+h-1)(x-1)h}=\lim_{h\to 0}\frac{-2}{(x+h-1)(x-1)}=-\frac{2}{(x-1)^2}.$$
   
   

   Another way is
   

   $$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{\frac{x+1}{x-1}-\frac{a+1}{a-1}}{x-a}$$
   
   
   
   
   $$=\lim_{x\to a}\frac{(x+1)(a-1)-(a+1)(x-1)}{(x-1)(a-1)(x-a)}=\lim_{x\to a}\frac{-2(x-a)}{(x-1)(a-1)(x-a)}=-\frac{2}{(a-1)^2}.$$
   
   

5. A ball is dropped from a tower of height $H$ feet. Its height after $t$ seconds, in feet above the ground, is given by the formula
   
   $$h(t) \ = \ H - 16t^2.$$
   
   
   a) (5) Find the velocity of the ball after $t$ seconds.

   
   

   The velocity function is the derivative of the height function with respect to the time $t$. Thus $v(t)=h'(t)=-32t$.

   
   

   b) (5) It is observed, that the ball hits the ground with a speed of $128$ feet per second. How long was the ball in the air? Hint: The speed is the absolute value of the velocity.

   
   

   At the moment when the ball hits the ground the velocity is $-128$ ft/s (the velocity vector points downward). We get $-32t=-128$, i.e. $t=4$, which means that the ball was 4 seconds in the air.

   
   

   c) (5) Use the information in part b) to determine the height of the tower.

   
   

   When the ball hits the ground the height of the ball is zero. On the other hand, it is the height of the ball after 4 seconds, thus we get $h(4)=0$, i.e. $H-16(4)^2=0$ and $H=256$.

6. a) (8) Suppose $a$ is a number, such that
   
   $$\lim_{h\rightarrow 0}\frac{a^h-1}{h} \ = \ 17.$$
   
   
   Find an equation for the tangent line to the graph of $y=a^x$ at the point $(0,1)$.

   
   

   The slope of the tangent line to the graph of $y=a^x$ at the point $x=0$ is equal to the derivative of $a^x$ at $x=0$. By definition the derivative of $a^x$ at $x=0$ is the limit ${\displaystyle \lim_{h\rightarrow 0}\frac{a^h-1}{h}}$. We are given that this limit is equal to 17, so the slope of the tangent is 17. Since the tangent passes through the point $(0,1)$ the equation of the tangent is $y-1=17(x-0)$, i.e. $y=17x+1$.

   
   

   b) (7) Find a formula for the derivative $f'(x)$ of the following function
   

   $$f(x) \ = \ x^{1984} + \frac{1}{x\sqrt{x}} - 2002 e^x.$$
   
   

   
   

   $$(x^{1984})'=1984x^{1983},\quad (2002e^x)'=2002e^x,$$
   
   

   
   

   $$\left(\frac{1}{x\sqrt{x}}\right)'=(x^{-1}x^{-1/2})'=(x^{-3/2})'=-3/2x^{-5/2}=-\frac{3}{2x^2\sqrt{x}},$$
   
   

   Therefore,
   

   $$f'(x)=1984x^{1983}-\frac{3}{2x^2\sqrt{x}}-2002e^x.$$
   
   

7. a) (10) Find a formula for the derivative $f'(x)$ of the function
   
   $$f(x) \ = \ \frac{x}{x-5}.$$
   
   

   By the quotient rule

   
   

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

   b) (5) Find the equation of the tangent line to the graph of $y=f(x)$ at the point $(4,-4)$.

   
   

   The slope of the tangent line to the graph of $y=f(x)$ at $(4,-4)$ is $f'(4)$. By part a)
   

   $$f'(4)=-\frac{5}{(4-5)^2}=-5.$$
   
   
   Since the line passes through $(4,-4)$ we have
   
   $$y+4=-5(x-4),\quad\rm { i.e. }\ \ y=-5x+16.$$