# Thread: Rate of change and limit of a function Unit questions

1. ## Rate of change and limit of a function Unit questions

Hi there, i have a few precalculus questions that im not sure how to solve, if you can help that would be great!

1. For what value of the constant c is the function f(x)= x + c, x < 2
cx(squared) + 1 > and equal 2
continuous at every number?

2. Can a limit exist at a(a as in variable) even though the function is discontinuous at a? Explain, use a graph

3. A man has been travelling in a spaceship. The height, in metres, as it approaches the moon is given by h= 10t(sqd) - 40t + 40 where t is in seconds. When and with what speed does it land on the moons surface?

Thanks alot hope you can help
sqd= squared

2. Originally Posted by mieniem
Hi there, i have a few precalculus questions that im not sure how to solve, if you can help that would be great!

1. For what value of the constant c is the function f(x)= x + c, x < 2
cx(squared) + 1 > and equal 2
continuous at every number?
You mean f(x)= x+ c for x< 2 and f(x)= $cx^2+ 1$ for x greater than or equal to x. what is the limit of f as x approaches 2 from below? ( $\lim_{x\to 2^-}f(x)= \lim_{x\to 2} x+ c$) What is the limit of f as x approaches 2 from above? ([tex]\lim_{x\to 2^+}f(x)= \lim_{x\to 2} cx^2+ 1[/maht])

What must be true for f to be continuous at x= 2?

2. Can a limit exist at a(a as in variable) even though the function is discontinuous at a? Explain, use a graph
What is the definition of "continuous" at a point? For a graph, draw any line and move one point.

3. A man has been travelling in a spaceship. The height, in metres, as it approaches the moon is given by h= 10t(sqd) - 40t + 40 where t is in seconds. When and with what speed does it land on the moons surface?

Thanks alot hope you can help
sqd= squared
Since h is the height at time t, it will reach the moon when h= 10t^2- 40t+ 40= 0. solve that for t.

Its speed is the derivative at that time. What is the derivative of t^2? Of 10t^2? What is the derivative of -40t?

For this particular problem, however, you do not need the derivative and, since this is in "precalculus" you may not be intended to do it that way. Notice that h= 10t^2- 40t+ 40= 10(t^2- 4t+ 4)= 10(T- 2)^2. h= 0 at the vertex of that parabola. What does the tangent line to the parabola look like at the vertex? The speed at that moment is the slope of the tangent line there.