b) (5) Support your result in part a) with a calculator. Describe your procedure and show as much data as possible.
As the values of approach 2 either from the right or from the left (but not equal to 2) the values of approach .
a) (5) Find and .
b) (5) What conditions do and have to satisfy, so that is continuous at .
For to be continuous at we need the limit exist and equal . Since and exists if and only if we see that the condition on and is .
c) (5) Find and , such that is everywhere differentiable.
First must be continuous everywhere, thus from part b). Notice that is differentiable at each . The derivative at is the limit . It exists if and only if the left hand and the right hand derivative limits are equal. We have
The vertical asymptotes are the lines such that the denominator is zero at , but the numerator is not. We have , i.e. or . Since and the vertical asymptotes are and .
The horizontal asymptotes are the lines where is the limit of as approaches positive or negative infinity. We have
Therefore the only horizontal asymptote is .
b) (5) Graph the function in part a) indicating all horizontal and vertical asymptotes.
One can first write . Then
Another way is
The velocity function is the derivative of the height function with respect to the time . Thus .
b) (5) It is observed, that the ball hits the ground with a speed of 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 ft/s (the velocity vector points downward). We get , i.e. , 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 , i.e. and .
The slope of the tangent line to the graph of at the point is equal to the derivative of at . By definition the derivative of at is the limit . 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 the equation of the tangent is , i.e. .
b) (7) Find a formula for the derivative of the following function
By the quotient rule
b) (5) Find the equation of the tangent line to the graph of at the point .
The slope of the tangent line to the graph of at
is . By part a)