Understanding the IVT, etc.

f(x) is continuous over [-2,2], f(-2)=-2 and f(2)=6. The Intermediate Value Theorem says that f(x) has at least one zero on [-2,2].

• True

• False

f(x) is continuous over [1,5], f(1)=4 and f(5)=4. The Intermediate Value Theorem says that f(x) has no zeros in the interval [1,5].

• True

• False

f(1)=2, f(10)=25. The Intermediate Value Theorem says that there is at least one value of x in the interval [1,10] where f(x)=17.

• True

• False

f(x) is continuous over the interval [-5,5], f(-5)<0 and f(5)<0. The Intermediate Value Theorem says that f(x) has no zeros on the interval [-5,5].

• True

• False

For the test, when is it ok to just evaluate a limit by plugging the point into the function?

• When the instructions don't say to use limit laws, a table of values, or a graph.

• When it works. (Sometimes it doesn't...)

• When the function is known to be continuous at the point.

• All of the above must be true.

How to we use the formal definition of continuity to show that f(x)=x+17 is continuous at the point a=3?

• Draw the graph and show that the function can be drawn without lifting your pencil.

• Show that f(x)=x+17 has no x values where it is undefined.

• Show that f(x)=x+17 is just a polynomial and therefore is continuous.

• Show that the limit as x approaches a of f(x) is the same as f(3).