You are here:

- Home
- Teens
- Homework/Study Tips
- Calculus
- Mathematics

Advertisement

Thank you for taking the time to answer my question. It is:

Determine the value of c so that the function below s continuous:

f(x)=

2x-5 x<2

cx^2 x(is equal to or greater than)2

That function is f(x) = { 2x - 5 for x < 2 / cx^2 for x>= 2 }.

To be continuous, the value at x=2 must be the same on both sides.

On the left it is 2x - 5 at x=2, and that is 2(2) - 5 = -1.

On the right, it is cx^2 for x=2, so it is 4c.

We need -1 = 4c, so solving gives us c = -1/4.

That way when we know the value is 4c, it is really -1, and that is the same.

Calculus

Answers by Expert:

Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology (reproduction, insusion of chemicals into bloodstream).

Experience in the area: I have tutored students in all areas of mathematics since 1980.
Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors.
Awards and Honors: I have passed Actuarial tests 100, 110, and 135.
**Publications**

Maybe not a publication, but I have respond to well oveer 8,500 questions on the PC.
Well over 2,000 of them have been in calculus.
**Education/Credentials**

I aquired well over 40 hours of upper division courses. This was well over the number that were required.
I graduated with honors in both my BS and MS degree from Oregon State University.
I was allowed to jump into a few courses at college a year early.
**Awards and Honors**

I have been nominated as the expert of the month several times.
All of my scores right now are at least a 9.8 average (out of 10).
**Past/Present Clients**

My past clients have been students at OSU, students at the college in South Seattle, referals from a company, friends and aquantenances, people from my church, and people like you from all over the world.