### Video Transcript

The figure shows the graph of the
function π on the closed interval from zero to 16 together with the dashed line
with equation π¦ equals 30. π of zero is less than 30 and π
of 16 is greater than 30. But π of π₯ is not equal to 30
anywhere on the closed interval from zero to 16. Why does this not violate the
intermediate value theorem?

Letβs just verify what weβre told
in the question. Is π of zero less than 30? Well, yes, we can see here π of
zero appears to be about 12. And likewise, π of 16 is greater
than 30. It appears to be about 32. But π of π₯ is not equal to 30
anywhere on the interval. This is true because the dashed
line π¦ equals 30 nowhere intersects the graph of our function. The question is, why does this not
to violate the intermediate value theorem? Well, what is the intermediate
value theorem? It states that if a function π is
continuous on a closed interval from π to π and if the number π is between π of
π and π of π, the values of the function on the end points of the interval. Then there exists π in the open
interval from π to π such that π of π is π.

We can see then how it might appear
that we have a counterexample to the intermediate value theorem. We set π equal to 30 and note that
30 is between π of zero and π of 16. However, there doesnβt exist any
value π in the open interval from zero to 16 such that π of π is 30. Why is this not a counterexample to
the intermediate value theorem? Well, the intermediate value
theorem only applies if π is continuous. Our function doesnβt satisfy this
required hypothesis. We can see that there is a
discontinuity here at π₯ equals eight.

So why does this not violate the
intermediate value theorem? Because the function is not
continuous at π₯ equals eight. And so, it is not continuous on the
closed interval from zero to 16 which would be required for the intermediate value
theorem to apply.