Intermediate Value Theorem 2 4 cont Examples

Intermediate Value Theorem 2. 4 cont.

Examples n If between 7 am and 2 pm the temperature went from 55 to 70. q q n At some time it reached 62. Time is continuous If between his 14 th and 15 th birthday, a boy went from 150 to 165 lbs. q q At some point he weighed 155 lbs. It may have occurred more than once.

Show that a “c” exists such that f(c)=2 for f(c)=x^2 +2 x-3 in the interval [0, 2] f(x) is continuous on the interval f(0)= -3 f(2)= 5

Determine if f(x) has any real roots f(x) is continuous on the interval f(1)= - # f(2)= + #

Is any real number exactly one less than its cube? (Note that this doesn’t ask what the number is, only if it exists. ) Since f is a continuous function, by the intermediate value theorem it must take on every value between -1 and 5. Therefore there must be at least one solution between 1 and 2. Use your calculator to find an approximate solution.

Max-Min Theorem for Continuous Functions If f is continuous at every point of the closed interval [a, b], then f takes on a minimum value m and a maximum value M on [a, b]. That is, there are numbers α and β in [a, b] such that f(α) = m, f(β) = M, and m ≤ f(x) ≤ M at all points x in [a, b] Max and mins at interior points Max and mins at endpoints Min at interior point and max at endpoint

Why does the IVT fail to hold for f(x) on [ -1, 1]? Not Continuous in interval! Point of discontinuity at x = 0

Show why a root exists in the given interval Continuous in interval f(-2)= -10 f(-1)= 1

Show why a root exists in the given interval Continuous in interval f(1)= 1